discrete random variables...chapter 10 • discrete random variables 467 2 list all the possible...

ChaPTer 10 • Discrete random variables 465

ChaPTer ConTenTS 10a Probability revision 10B Discrete random variables 10C Measures of centre of discrete random distributions 10d Measures of variability of discrete random distributions

ChaPTer 10

Discrete random variables

diGiTal doCdoc-922110 Quick Questions

10a Probability revisionTo introduce this chapter we shall revise important concepts and skills that were covered in Mathematical Methods (CAS) Units 1 and 2.

TerminologyThe circular spinner at right is divided into 8 equal sectors. When the spinner is spun, the possible outcomes are 1, 2, 3, 4, 5, 6, 7, 8. These outcomes may be listed as the elements of a set. The set of all possible outcomes of an experiment is called the sample space (or the universal set) and is denoted by ε, and each possible outcome is called a sample point. Therefore, spinning the spinner gives ε = {1, 2, 3, 4, 5, 6, 7, 8}.

A subset of the sample space is known as an event. For the example above, if event A is defi ned as ‘an odd number when the spinner is spun’, then A = {1, 3, 5, 7}. If event B is defi ned as ‘a number less than 5 when the spinner is spun’, then B = {1, 2, 3, 4}.

The union (symbol ∪) of the two events A and B above implies a combined event, that is, either event A or event B or both occurring. Therefore the set A ∪ B = {1, 2, 3, 4, 5, 7}. Note: Common elements are written only once.

The intersection (symbol ∩) of the two events A and B above is represented by the common sample points of the two events. Therefore the set A ∩ B = {1, 3}.

Venn diagramsVenn diagrams involve drawing a rectangle that represents the sample space and a series of circles that represent subsets of the sample space. They provide a visual representation of the information at hand and clearly display the relationships between sets.

The Venn diagrams below illustrate an alternative way of presenting information regarding the circular spinner shown on the previous page.

86

A B5

31 2

7 4

A ∪ B

ε

86

A B5

31 2

7 4

A ∩ B

ε

A

86

A B5

31 2

7 4

ε

Concept summary

Read a summary of this concept.

Units: 3 & 4

AOS: 4

Topic: 1

Concept: 11 2

3

4

56

7

8

ConTenTS

10a Probability revision 465 exercise 10a Probability revision 477

10B discrete random variables 479 exercise 10B discrete random variables 484

10C measures of centre of discrete random distributions 487 exercise 10C measures of centre of discrete random distributions 492

10d measures of variability of discrete random distributions 494 exercise 10d measures of variability of discrete random distributions 501

466 Maths Quest 12 Mathematical Methods CAS

Note: Sample points not belonging to either set are placed outside the circles but remain inside the rectangle.

Venn diagrams can also assist in determining whether or not two sets are equal, that is, whether they contain the same elements. As the examples below show, the equality of two sets may not be obvious from the set notation but is often easier to see in a diagram.

A B

A′ ∩ B′ = (A ∪ B)′

A B

A′ ∪ B′ = (A ∩ B)′

ProbabilityProbability deals with the likelihood or chance of some event occurring. The probability of a specific event, say A, occurring is defined by the rule:

Pr(A) = number of favourable outcomes .

total number of possible outcomes

Its probability lies within the restricted interval 0 ≤ Pr(A) ≤ 1. A probability of zero implies that the event cannot occur, while a probability of 1 implies that the event will most certainly occur.

The individual probabilities of a particular experiment will sum to a value of 1 and can be denoted as follows.

p x( ) 1∑ =

If Pr(A) is defined as the probability of an event occurring, then its complement, Pr(A′), is defined as the probability of an event not occurring.Therefore, it can be stated that Pr(A) + Pr(A′) = 1which can be transposed to Pr(A′) = 1 − Pr(A).

Worked examPle 1

Two fair dice are rolled simultaneously and the sum of the two numbers appearing uppermost is recorded as shown below.

(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

Find the probability that the sum will be:a 6 b 10 c a number less than 5 d at least 9 e an odd number.

Think WriTe

a 1 Define the event. a (5, 1) (4, 2) (3, 3) (2, 4) (1, 5)

2 List all the possible favourable outcomes. Let A = the sum of 6.

3 Substitute the values into the probability rule. Pr(A) = number of favourable outcomes


= 536

b 1 Define the event. b (6, 4) (5, 5) (4, 6)


2 List all the possible favourable outcomes. Let A = the sum of 10.



= 3

36

4 Simplify.1

12=

c 1 Defi ne the event. c (1, 1) (2, 1) (3, 1) (1, 2) (2, 2) (1, 3)

2 List all the possible favourable outcomes. Let A = a number less than 5.


total number of possible outcomes6

36=

4 Simplify. 1

6=

d 1 Defi ne the event. d (6, 3) (5, 4) (6, 4) (4, 5) (5, 5)(6, 5) (3, 6) (4, 6) (5, 6) (6, 6)

2 List all the possible favourable outcomes. Let A = at least 9.



36=

4 Simplify. 5

18=

e 1 Defi ne the event. e (2, 1) (4, 1) (6, 1) (1, 2) (3, 2) (5, 2)(2, 3) (4, 3) (6, 3) (1, 4) (3, 4) (5, 4)(2, 5) (4, 5) (6, 5) (1, 6) (3, 6) (5, 6)

2 List all the possible favourable outcomes. Let A = an odd number.



36=

4 Simplify. 1

2=

Worked examPle 2

A bag contains 15 marbles comprising 5 black, 3 red, 4 blue, 2 white and 1 green. One marble is drawn randomly from the bag.a Determine the probability of each of the coloured marbles being drawn: i black ii red iii blue iv white v green.b Show that the probabilities sum to 1.c What is the probability that the marble drawn is: i not black? ii either black or white? iii neither blue nor green?

Think WriTe

a i 1 Defi ne the event. a i Let B = a black marble.

2 Substitute the values into the probability rule.

Pr(B) = number of favourable outcomes


= 5

15


3 Simplify. = 13

ii 1 Defi ne the event. ii Let R = a red marble.


Pr(R) = number of favourable outcomes


= 3

15

3 Simplify. = 15

iii 1 Defi ne the event. iii Let Bl = a blue marble.


Pr(Bl) = number of favourable outcomes


= 4

15

iv 1 Defi ne the event. iv Let W = a white marble.


Pr(W) = number of favourable outcomes


= 2

15

v 1 Defi ne the event. v Let G = a green marble.


Pr(G) = number of favourable outcomes


= 1

15

b Add each of the probabilities. b Sum of probabilities = 13 + 1

5 + 4

15 + 2

15 + 1

15

= 1

c i 1 Write the appropriate rule: Pr(A′) = 1 − Pr(A).

c i Pr(B′) = 1 − Pr(B)

2 Substitute the known values into the rule. = 1 − 13

3 Evaluate. = 23

ii 1 Add each of the probabilities together. ii Pr(B ∪ W) = Pr(B) + Pr(W)

2 Substitute the known values into the rule. = 13 + 2

15

3 Evaluate. = 715

iii 1 Write the appropriate rule. iii Pr(Bl′ ∪ G′) = 1 − [Pr(Bl) + Pr(G)]

2 Substitute the known values into the rule. = 1 4

15

1

15( )− +

3 Evaluate. = 1 − 515

= 1015

4 Simplify. = 23


The addition rule of probabilityThe addition rule of probability states that Pr(A ∪ B) = Pr(A) + Pr(B) − Pr(A ∩ B).

Worked examPle 3

a If Pr(A) = 0.4, Pr(B) = 0.7 and Pr(A ∩ B) = 0.2, find Pr(A ∪ B).b If Pr(A) = 0.6, Pr(B) = 0.8 and Pr(A ∪ B) = 0.9, find Pr(A ∩ B).

Think WriTe

a 1 Write the addition rule. a Pr(A ∪ B) = Pr(A) + Pr(B) − Pr(A ∩ B)

2 Substitute the known values into the rule. = 0.4 + 0.7 − 0.2

3 Evaluate. = 1.1 − 0.2= 0.9

b 1 Write the addition rule. b Pr(A ∪ B) = Pr(A) + Pr(B) − Pr(A ∩ B)

2 Substitute the known values into the rule. 0.9 = 0.6 + 0.8 − Pr(A ∩ B)0.9 = 1.4 − Pr(A ∩ B)

3 Transpose the equation to make Pr(A ∩ B) the subject.

Pr(A ∩ B) = 1.4 − 0.9

5 Evaluate. = 0.5

Venn diagrams may also be used to display the probabilities rather than just the outcomes, as shown in the diagram below.

Pr (A′∩B′)

A B

Pr (A∩B′) Pr (A′∩B)

Pr (A∩B)

ε

Note: Pr(A ∪ B) is represented by the shaded section.The probabilities given and calculated in Worked example 3(a) and 3(b) can be displayed as follows.

Worked example 3(a)

A B

0.2 0.2 0.5

0.1

ε

Worked example 3(b)

A B

0.1 0.5 0.3

0.1

ε

mutually exclusive eventsIf two or more events cannot occur simultaneously, they are said to be mutually exclusive or disjoint; that is, they have nothing in common. In set notation this may be expressed as Pr(A ∩ B) = { } or Pr(A ∩ B) = 0.

Therefore, for mutually exclusive events the addition rule becomes:

Pr(A ∪ B) = Pr(A) + Pr(B).

independent eventsTwo events A and B are independent if one event does not influence the other event from occurring. The mathematical definition of independence is given by:

Pr(A ∩ B) = Pr(A) × Pr(B).


Worked examPle 4

Two fair dice are rolled with S representing the event of obtaining a number less than 4 on the first die and T the event of obtaining a number greater than 4 on the second die. Find:a Pr(S)b Pr(T)c if events S and T are mutually exclusived if events S and T are independent.

Think WriTe

a 1 Refer to the dice results recorded in the question in Worked example 1.

a

2 Define the event. Let S = obtaining a number less than 4 on the first die.

3 Determine the probability. Pr(S) = 1836

4 Simplify. = 12

b 1 Define the event. b Let T = obtaining a number greater than 4 on the first die.

2 Determine the probability. Pr(T) = 1236

3 Simplify. = 13

c Answer the question with reasoning. c Events S and T are not mutually exclusive since they have common points; that is, (1, 5) (1, 6) (2, 5) (2, 6) (3, 5) (3, 6).

d 1 Answer the question using the dice results in Worked example 1.

d From the dice results recorded in the question in Worked example 1,Pr(S ∩ T) = 6

36

= 16

2 Check with answer obtained using the rule. Using the rule Pr(S ∩ T) = Pr(S) × Pr(T)

= 12 ×

1

3

= 16

Since both methods give the same answer, S and T are independent events.

Worked examPle 5

Two fair dice are rolled with U representing the event of obtaining a 5 on the first die and V the event of the sum of numbers on the two dice exceeding 10. Find:a Pr(U) b Pr(V) c if events U and V are independent.

Think WriTe

a 1 Refer to the dice results recorded in the question in Worked example 1.

a

2 Define the event. Let U = obtaining a 5 on the first die.

3 Determine the probability. Pr(U) = 636

4 Simplify. = 1

6

b 1 Define the event. b Let V = the sum of the numbers on the two dice exceeds 10.


2 Determine the probability. Pr(V) = 3

36

3 Simplify. = 1

12

c 1 Answer the question using the dice results. c From the dice results, Pr(U ∩ V) = 136

2 Check with answer obtained using the rule. Using rule Pr(U ∩ V) = Pr(U) × Pr(V)

= 16 × 1

12

= 172

Since the two methods do not give the same answer, U and V are not independent events.

karnaugh maps and probability tablesKarnaugh maps and probability tables summarise all combinations of two events (for example A and B) and their complements (for example A′ and B′ ).

B B′A A ∩ B A ∩ B′A′ A′ ∩ B A′ ∩ B′

B B′A Pr(A ∩ B) Pr(A ∩ B′) Pr(A)

A′ Pr(A′ ∩ B) Pr(A′ ∩ B′) Pr(A′)Pr(B) Pr(B′) 1

Karnaugh map Probability table

Worked examPle 6

For the probability table shown at right, event A is ‘not more than 17 years of age’ and event B is ‘has a learner permit’.a Complete the probability table at right.b What do the following probabilities

represent? i Pr(A ∩ B) ii Pr(A′ ∩ B′)c What is the probability that: i a person over the age of 17 does not have a learner permit? ii a person has a learner permit and is older than 17? iii a person over the age of 17 has a learner permit or a person at or under the age of 17 does not

have a learner permit?

Think WriTe

a 1 Calculate the values of the cells in row 1, column 3 and row 3, column 1.

a Pr(A) = 1 − Pr(A′) Pr(B) = 1 − Pr(B′)= 1 − 0.5 = 1 − 0.63= 0.5 = 0.37

2 Enter the values into the appropriate cells.

Column 1 Column 2 Column 3

B B′Row 1 A 0.12 0.5

Row 2 A′ 0.5

Row 3 0.37 0.63 1

3 Calculate the value of the cell in row 1, column 2.

Pr(A ∩ B′) = Pr(A) − Pr(A ∩ B)= 0.5 − 0.12= 0.38


B B′Row 1 A 0.12

Row 2 A′ 0.5

Row 3 0.63 1


4 Enter the value into the appropriate cell.


B B′Row 1 A 0.12 0.38 0.5

Row 2 A′ 0.5

Row 3 0.37 0.63 1


Pr(A′ ∩ B) = Pr(B) − Pr(A ∩ B) = 0.37 − 0.12 = 0.25



B B′Row 1 A 0.12 0.38 0.5

Row 2 A′ 0.25 0.5

Row 3 0.37 0.63 1


Pr(A′ ∩ B′) = Pr(B′) − Pr(A ∩ B′) = 0.63 − 0.38 = 0.25



B B′Row 1 A 0.12 0.38 0.5

Row 2 A′ 0.25 0.25 0.5

Row 3 0.37 0.63 1

b i Explain what Pr(A ∩ B) represents in this example.

b i Pr(A ∩ B) represents the probability of a person at or under the age of 17 having a learner permit. In this case the probability of the given event occurring is 0.12.

ii Explain what Pr(A′ ∩ B′) represents in this example.

ii Pr(A′ ∩ B′) represents the probability of a person over the age of 17 not having a learner permit. In this case the probability of the given event occurring is 0.25.

c i State the appropriate probability from the table.

i Pr(A′ ∩ B′) = 0.25

ii State the appropriate probability from the table.

ii Pr(A′ ∩ B) = 0.25

iii State the appropriate probabilities from the table and evaluate.

iii Pr(A′ ∩ B) + Pr(A ∩ B′) = 0.25 + 0.38 = 0.63

Conditional probabilityConditional probability deals with an event which has previously occurred and has an effect on an event we are interested in. Due to the initial condition (or restriction) imposed, the number of possible events

is reduced. Conditional probability is defi ned by the rule A BA B

BPr( )

Pr( )

Pr( ),| = ∩ where Pr(B) ≠ 0, and

can be transposed to Pr(A ∩ B) = Pr(A | B) × Pr(B). The latter is called the multiplication rule. Pr(A | B) is read as ‘the probability of A given B’.

Concept summary


Units: 3 & 4

AOS: 4

Topic: 1

Concept: 2


Worked examPle 7

If Pr(A) = 1155, Pr(B) = 11

1100 and Pr(A ∩ B) = 11

2200, fi nd:

a Pr(A ∪ B) b Pr(A | B) c Pr(B | A)d if events A and B are mutually exclusivee if events A and B are independent.

Think WriTe

a 1 Write the addition rule. a Pr(A ∪ B) = Pr(A) + Pr(B) − Pr(A ∩ B)

2 Substitute the known values into the rule. = 15 + 110 − 120

3 Evaluate. = 520

4 Simplify. = 14

b 1 Write the appropriate rule. b Pr(A | B) = A B

B

Pr( )

Pr( )

∩

2 Substitute the known values into the rule. = 120110

3 Evaluate. = 120 ÷ 110

= 120 × 101

= 1020

4 Simplify. = 12

c 1 Write the appropriate rule. c Pr(B | A) = B A

A

Pr( )

Pr( )

∩


3 Evaluate. = 120 ÷ 15

= 120 × 51

= 520

4 Simplify. = 14

d State the answer, showing reasoning. d Events A and B are not mutually exclusive since they have common events, that is,

Pr(A ∩ B) = 120

.

e Compare the given value with the answer obtained using the rule.

e Pr(A ∩ B) = 120 .

Using rule Pr(A ∩ B) = Pr(A) × Pr(B)

= 15 × 110

= 150

Since the two methods do not give the same answer, A and B are not independent events.

From Worked example 7(b) and 7(c) it can be seen that Pr(A | B) ≠ Pr(B | A).

Tree diagramsTree diagrams are a useful tool in solving probability tasks as they display each of the possible outcomes along with their respective probabilities.

TUTorialeles-1226Worked example 7


Worked examPle 8

Nadia knows that if her car starts, she has an 80% chance of getting to work on time. However, if her car doesn’t start, her chance of arriving on time is 50%. If Nadia’s car starts only 70% of the time, what is the probability that:a her car starts and she gets to work on time?b she arrives at work late?c she arrives at work on time?d her car starts, given that she arrives at work on

time?

Think WriTe/draW

a 1 Define the events. a Let C = car startsLet C′ = car doesn’t startLet O = Nadia arrives at work on timeLet L = Nadia arrives at work late

2 Assign probabilities to each event.

'

C

C

Pr( )

Pr( )

70100710310

=

=

=

If car starts, Pr(O) = 80

100

= 4

5

If car starts, Pr(L) = 1

5

If car doesn’t start, Pr(O) = 50

100

= 1

2

If car doesn’t start, Pr(L) = 1

2

3 Draw a tree diagram with each branch assigned the appropriate probability. 7–

10

1–2

1–2

4–5

1–5

3–10

CO

L

LC′

O

4 Calculate the required probability. Pr(CO) = 710

× 45

= 2850

5 Simplify. = 1425

b Calculate the required probability. b Pr(L) = Pr(CL) + Pr(C′L)

= 710

× 15 + 3

10 × 1

2

= 750

+ 320

= 14100

+ 15100

= 29100


c Calculate the required probability. c Pr(O) = Pr(CO) + Pr(C′O)

= 7

10 × 45 + 3

10 × 1

2

= 2850

+ 320

= 56100

+ 15100

= 71100

d 1 Write the appropriate rule.Note: ‘Given’ implies conditional probability.

d Pr(C | O) = CO

O

Pr( )

Pr( )


3 Evaluate and simplify. = 1425

71100

÷

= 1425

10071×

= 5671

Worked examPle 9

A fair coin is tossed three times. Find the probability of obtaining two heads given the first toss resulted in a tail.

Think WriTe/draW

1 Draw a tree diagram and list all of the possible outcomes.

H

TH

TT

H

TH

H

TT

H

T

H

HHH, HHT, HTH, HTT, THH, THT, TTH, TTT

2 Write the appropriate rule. Pr(2H | tail first toss) = Pr(2H tail first toss)

Pr(tail first toss)

∩

3 Calculate the probability of each event. Pr(2H ∩ tail first toss) = 18

Pr(tail first toss) = 48

= 12

4 Substitute the known values into the rule. Pr(2H | tail first toss) =

1812

5 Evaluate. = 18

12÷

= 18 × 2

1

= 28

6 Simplify. = 14


CombinationsIn mathematics, a combination deals with the number of ways items may be selected from a set of elements where the order is not important. For example, in how many ways can 3 numbers be selected from the set {1, 2, 3, 4}, taking into account that order is not important?

The following selections can be made:1, 2, 3 1, 2, 4 2, 3, 4 3, 4, 1

Hence 4 selections could be made.If order was important, there would be a greater number of possibilities since each of the above

selections could be arranged in 6 ways. For example, the selection (1, 2, 3) could be arranged as:1, 2, 3 2, 1, 3 2, 3, 1 3, 1, 2 3, 2, 1 1, 3, 2.

A combination is also referred to as a selection or choice, and is defi ned by the rule nCr.nCr = the number of selections of n different objects taken r at a time

n

n r r

!

! !( )=

−

nCr may also be expressed as rn( ) and is read as ‘n over (above) r’.

Worked examPle 10

A drawer contains 7 T-shirts of which 3 are white and the rest are black. If 2 T-shirts are randomly selected from the drawer simultaneously, fi nd the probability that they are:a both black b both whitec different colours d the same colour.

Think WriTe

a 1 Calculate the number of selections of taking 2 black T-shirts from a total of 4.

a 4C2 = 6; that is, there are 6 ways of selecting 2 black T-shirts from a total of 4.

2 Calculate the number of selections of taking 2 T-shirts from a total of 7.

7C2 = 21; that is, there are 21 ways of selecting 2 T-shirts from a total of 7.

3 Calculate the probability using the rule. Pr(both black) =2 black T-shirts from 4

2 T-shirts from 7


21

5 Simplify. = 27

b 1 Calculate the number of selections of taking 2 white T-shirts from a total of 3.

b 3C2 = 3; that is, there are 3 ways of selecting 2 white T-shirts from a total of 3.



3 Calculate the probability using the rule. Pr(both white) =2 white T-shirts from 3

2 T-shirts from 7


21

5 Simplify. = 17

c 1 Calculate the number of selections of taking 1 black T-shirt from a total of 4.

c 4C1 = 4; that is, there are 4 ways of selecting 1 black T-shirt from a total of 4.

2 Calculate the number of selections of taking 1 white T-shirt from a total of 3.

3C1 = 3; that is, there are 3 ways of selecting 1 white T-shirt from a total of 3.





4 Calculate the probability using the rule. Pr(different colours) =1 black × 1 white

2 T-shirts from 7

5 Substitute the known values into the rule. = 4 3

21

×

6 Evaluate. = 1221

7 Simplify. = 47

d 1 Calculate the probability using the rule. d Pr(same colours) = Pr(both black) + Pr(both white)

2 Substitute the known values into the rule. = 27 + 1

7

3 Evaluate. = 37

exercise 10a Probability revision1 We1 Two fair dice are rolled simultaneously and the sum of the two numbers appearing uppermost is

recorded. Find the probability that the sum will be:a 3 b 12 c 7 d greater than 4e at least 7 f an even number g a prime number.

2 We2 A bag contains 12 marbles comprising 3 black, 5 red and 4 green. One marble is drawn randomly from the bag.a Determine the probability of each of the coloured marbles being drawn:

i black ii red iii green.b Show that the probabilities sum to 1.c What is the probability that the marble drawn is:

i not green? ii either black or red?iii neither red nor green? iv either black, red or green?

3 A fair coin is tossed three times. Find the probability of obtaining:a three heads b two headsc one head d no headse at least two heads.

4 A circular spinner is divided into 8 equal sectors and numbered as shown in the diagram at right. If the spinner is spun once, find the probability of obtaining:a a one b a two c a three or a four or a fived a one or a two.

5 We3a If Pr(A) = 0.3, Pr(B) = 0.6 and Pr(A ∩ B) = 0.2, find Pr(A ∪ B).

6 We3b If Pr(A) = 0.5, Pr(B) = 0.4 and Pr(A ∪ B) = 0.8, find Pr(A ∩ B).

7 If Pr(A) = 4 Pr(B), Pr(A ∪ B) = 0.8 and Pr(A ∩ B) = 0.2, find: a Pr(B) b Pr(A).

8 Of the 200 students studying VCE at Merlynston Secondary College, 80 study Maths Methods, while there are 65 Physics students. If there are 85 students who don’t take either Maths Methods or Physics, find the probability that a randomly selected student:a studies Maths Methods b studies Physicsc studies neither Maths Methods nor Physics d studies Maths Methods and Physicse studies Physics, given that the student studies Maths Methods.

9 We4,5 Two fair dice are rolled, with F representing the event of obtaining a number greater than 4 on the first die and G the event of obtaining an even number on the second. Find:a Pr(F) b Pr(G)c if events F and G are mutually exclusived if events F and G are independent.

10 For two events P and Q, Pr(P) = 0.72, Pr(Q) = 0.25 and Pr(P ∪ Q) = 0.91. Are P and Q mutually exclusive events?

11 For two events X and Y, Pr(X) = 0.4, Pr(Y) = 0.5 and Pr(X ∩ Y) = 0.2. Are X and Y independent events?

1 1

2

2

34

5

1


12 We6 For the probability table shown, A is the event ‘is unfit’ and B is the event ‘is a smoker’.a Complete the probability table at right.b What do the following probabilities

represent? i Pr(A ∩ B) ii Pr(A′ ∩ B′)c What is the probability that: i a person is unfi t and a non-smoker? ii a person is a smoker and fi t? iii a person is unfi t and a smoker or is

unfi t and a non-smoker? iv a person is a non-smoker?

13 We7 If Pr(A) = 12 , Pr(B) = 13 and Pr(A ∩ B) = 16 find:

a Pr(A ∪ B) b Pr(A | B) c Pr(B | A)d if events A and B are mutually exclusivee if events A and B are independent.

14 If Pr(A) = 0.4, Pr(B) = 0.5 and Pr(A ∩ B) = 0.2 find:

a Pr(A ∪ B) b Pr(A | B) c Pr(B | A).

15 We8 A recent study has shown that 60% of people who don’t wear glasses get regular headaches, while only 30% of people who wear glasses are headache sufferers. If 35% of people wear glasses, find the probability that a randomly selected person:a wears glasses and gets headachesb does not wear glasses and suffers from headachesc suffers from headachesd wears glasses, given that the person suffers from headaches.

16 Jemma knows that if her alarm goes off, she has a 90% chance of getting to school on time. However, if the alarm does not ring, her chance of arriving on time is only 40%. If Jemma’s alarm clock works only 60% of the time, what is the probability that:a she gets to school on timeb she arrives late to schoolc her alarm rang, given that she arrived on time?

17 A bag contains 5 red marbles and 3 green marbles. A marble is selected at random, its colour is observed and it is then replaced. A second selection is then made. Find the probability that the two marbles chosen were:a both red b both greenc different colours d the same colour.

18 mC Two fair dice are rolled. The probability of the numbers showing uppermost on both dice being the same is:

a 1

36 B 1

18 C 1

6 d

1

3 e 1

2

19 mC If Pr(S) = 0.2, Pr(T) = 0.5 and Pr(S ∪ T) = 0.6, which one of the following is not true?a Pr(S ∩ T) = 0.1 B Pr(S | T) = 0.2 C Pr(T | S) = 0.5d S and T are mutually exclusive. e S and T are independent.

20 mC The probability of picking a red picture card from a standard pack of playing cards is:

a 1

2 B 3

13 C 2

13 d 3

26 e 1

26

21 mC If Pr(M) = 0.3, Pr(N) = 0.4 and Pr(M | N) = 0.5 then Pr(M ∩ N) is equal to:a 0.15 B 0.2 C 0.6 d 0.75 e 0.8

22 We9 A fair coin is tossed three times. Find the probability of obtaining three tails, given that the first toss resulted in a tail.

23 If Pr(A) = 0.6, Pr(B) = 0.5 and Pr(A ∩ B) = 0.36, find:a Pr(A′) b Pr(B′) c Pr(A ∪ B)d Pr(A′ ∩ B′) e Pr(A′ ∪ B) f Pr(A | B)g Pr(B | A) h Pr(A | B′) i Pr(B | A′).


B B′

Row 1 A 0.22

Row 2 A′ 0.60 0.68

Row 3 1


24 We10 A drawer contains 6 T-shirts, of which 2 are white and the rest are black. If 2 T-shirts are randomly selected from the drawer simultaneously, find the probability that they are:a both black b both white c different colours d the same colour.

25 A box contains one dozen chocolates, of which 4 are strawberry creams, 3 are orange creams and 5 are peppermint creams. Two chocolates are selected at random. Find the probability that they are both the same type if:a the fi rst chocolate is replaced before the second is drawnb the fi rst chocolate is not replaced before the second is drawn.

26 mC A fair die has its 4-spot changed to a 5-spot and its 2-spot changed to a 3-spot. The probability of getting an even number when the altered die is rolled is:

a 1

6B 1

3C 1

2d 2

3e 5

6

27 mC A box contains 3 red balls and 2 green balls. Two balls are chosen simultaneously. The probability that they are the same colour is:

a 8

25B 2

5C 13

25d 3

5e 13

20

28 A bag contains 5 red cubes and 3 black cubes. Three cubes are chosen at random. Find the probability of at least 2 reds being chosen, given that the first cube was red:a if the cubes are replaced after each drawb if the cubes are not replaced after each draw.

29 Joanne knows that her chance of winning each tennis match she plays is 0.8. A knockout tournament requires players to win five matches to win the championship. What is the probability that Joanne:a wins the tournament? Give your answer to 4 decimal places.b wins the tournament given that she wins her fi rst three matches?

30 In a particular suburb the chances of a woman owning her own home is 0.4, while the probability of a woman owning her own home and being employed is 0.2. Find the probability that a woman who owns her own home is also employed.

31 The probability of Vanessa’s car starting on a cold morning is 0.6, while on a normal morning the chance of it starting is 0.9. The probability of any morning being a cold one is 0.3. If Vanessa’s car starts tomorrow morning, find the probability that the morning is cold.

32 The Roosters know that they will win 80% of their home matches and 40% of their away matches. This season’s fixture has the Roosters playing 55% of their games at home. Given that the Roosters won their last game, what was the probability that it was played at home?

33 Tatiana is trying out for a place on the high jump team. In order to qualify she must clear three of the four heights. She knows that she has a 70% chance of clearing the first height and a 65% chance of clearing any subsequent height. What is the probability, to 4 decimal places, that Tatiana:a clears the fi rst, third and fourth heights only?b clears three heights?c clears three heights, given she did not clear the fi rst height?

10B discrete random variablesA random variable is one whose value cannot be predicted but is determined by the outcome of an experiment. For example, two dice are rolled simultaneously a number of times. The sum of the numbers appearing uppermost is recorded. The possible outcomes we could expect are {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}. Since the possible outcomes may vary each time the dice are rolled, the sum of the numbers appearing uppermost is a random variable.

Random variables are expressed as capital letters, usually from the end of the alphabet (for example, X, Y, Z) and the value they can take on is represented by lowercase letters (for example, x, y, z respectively).

The above situation illustrates an example of a discrete random variable since the possible outcomes were able to be counted. Discrete random variables generally deal with number or size.

A random variable that can take on any value is defi ned as a continuous random variable. Continuous random variables generally deal with quantities that can be measured, such as mass, height or time.


Worked examPle 11

Which of the following represent discrete random variables?a The number of goals scored at a football matchb The height of students in a Maths Methods classc Shoe sizes d The number of girls in a fi ve-child familye The time taken, in minutes, to run a distance of 10 kilometres

Think WriTe

Determine whether the variable can be counted or needs to be measured.

a Goals can be counted. a Discrete

b Height must be measured. b Continuous

c The number of shoe sizes can be counted. c Discrete

d The number of girls can be counted. d Discrete

e Time must be measured. e Continuous

discrete probability distributionsWhen dealing with random variables, the probabilities associated with them are often required.

Worked examPle 12

Let X represent the number of tails obtained in three tosses. Draw up a table that displays the values the discrete random variable can assume and the corresponding probabilities.

Think WriTe/draW

1 Draw a tree diagram and list all of the possible outcomes.

H

TH

TT

H

TH

H

TT

H

T

H

HHH, HHT, HTH, HTT, THH, THT, TTH, TTT

2 Draw a table with two columns: one labelled number of tails, the other probability.

Number of tails (x) Probability Pr(x)

0 1

8

1 3

8

2 3

8

3 1

8

3 Enter the information into the table.

The table above displays the probability distribution of the total number of tails obtained in three tosses of a fair coin. Since the variable in this case is discrete, the table displays a discrete probability distribution.

In Worked example 12, X denoted the random variable and x the value that the random variable could take. Thus the probability can be denoted by p(x) or Pr (X = x). Hence the table in Worked example 12 could be presented as shown below.

x 0 1 2 3

Pr(X = x) 18

3

838

18

Concept summary


Units: 3 & 4

AOS: 4

Topic: 1

Concept: 3

See more Watch

a video about discrete probability distributions.

Concept summary


Units: 3 & 4

AOS: 4

Topic: 1

Concept: 4


Close inspection of this table shows important characteristics that satisfy all discrete probability distributions.1. Each probability lies in a restricted interval 0 ≤ Pr(X = x) ≤ 1.2. The probabilities of a particular experiment sum to 1, that is,

x xPr 1( )∑ ===

If these two characteristics are not satisfi ed, then there is no discrete probability distribution.

Worked examPle 13

Draw a probability distribution graph of the outcomes in worked example 12.

Think WriTe/draW

1 Draw a set of axes in the fi rst quadrant only. Label the horizontal axis x and the vertical axis Pr(X = x).

x0 1 2 3

3–8

2–8

1–8

Pr(X = x)

2 Mark graduations evenly along the horizontal and vertical axes, and label with appropriate values.

3 Draw a straight line from each x-value to its corresponding probability.

Note: The probability distribution graph may also be drawn as follows.

x0 1 2 3

3–8

2–8

1–8

Pr(X = x)

x0 1 2 3

3–8

2–8

1–8

Pr(X = x)

A column graph A dot graph

Worked examPle 14

Which of the following tables represent a discrete probability distribution?a x 0 1 2 3

Pr(X = x) 0.2 0.5 0.2 0.1

b x 0 2 4 6

Pr(X = x) 0.5 0.3 0.1 0.1

c x −1 0 1 2

Pr(X = x) 0.2 0.1 0.3 0.3

d x −2 0 5 7

Pr(X = x) −0.2 0.3 0.5 0.4

Think WriTe

a 1 Check whether each of the probabilities lie within the restricted interval 0 ≤ Pr(X = x) ≤ 1.

a All probabilities lie between 0 and 1 inclusive.

2 Check that the probabilities sum to 1. 0.2 + 0.5 + 0.2 + 0.1 = 1

3 Answer the question. Yes, this is a discrete probability distribution since both requirements have been met.

b 1 Check whether each of the probabilities lie within the restricted interval 0 ≤ Pr(X = x) ≤ 1.

b All probabilities lie between 0 and 1 inclusive.

2 Check that the probabilities sum to 1. 0.5 + 0.3 + 0.1 + 0.1 = 1


3 Answer the question. Yes, this is a discrete probability distribution since both requirements have been met.

c 1 Check whether each of the probabilities lie within the restricted interval 0 ≤ Pr(X = x) ≤ 1.

c All probabilities lie between 0 and 1 inclusive.

2 Check that the probabilities sum to 1. 0.2 + 0.1 + 0.3 + 0.3 ≠ 1 (totals to 0.9)

3 Answer the question. No, this is not a discrete probability distribution since both requirements have not been met.

d 1 Check whether each of the probabilities lie within the restricted interval 0 ≤ Pr(X = x) ≤ 1.

d The fi rst probability is a negative value, so not all probabilities lie between 0 and 1 inclusive.

2 Check that the probabilities sum to 1. −0.2 + 0.3 + 0.5 + 0.4 = 1

3 Answer the question. No, this is not a discrete probability distribution since both requirements have not been met.

Worked examPle 15

Find the value of k for each of the following discrete probability distributions.a x 1 3 5 7 9

Pr(X = x) 0.2 k 0.2 0.3 0.1

b x 0 1 2 3 4

Pr(X = x) 5k 6k 4k 3k 2k

Think WriTe

a 1 Add up each of the given probabilities.They should sum to 1.

a X xPr( ) 1∑ = =0.2 + k + 0.2 + 0.3 + 0.1 = 1 0.8 + k = 1 k = 1 − 0.8 = 0.2

2 Simplify.

3 Solve to fi nd k.

b 1 Add up each of the given probabilities. They should sum to 1.

b 5k + 6k + 4k + 3k + 2k = 1

2 Simplify. 20k = 1

3 Solve to fi nd k. k = 120

Worked examPle 16

a Show that the function p(x) = 142

(5x + 3), where x = 0, 1, 2, 3 is a probability function.

b Show that the function p(x) = 1

100 x2 (6 − x), where x = 2, 3, 4, 5 is a

probability function.

Think WriTe

a 1 Substitute each of the x-values into the equation and obtain the corresponding probability.

a When x = 0, p(x) = 342

= 114



2 Simplify where possible. When x = 1, p(x) = 842

= 421

When x = 2, p(x) = 1342

When x = 3, p(x) = 1842

= 37

3 Check whether each of the probabilities lie within the restricted interval 0 ≤ Pr(X = x) ≤ 1.

All probabilities lie between 0 and 1 inclusive.

4 Check whether the probabilities sum to 1. 1

14+ 4

21 + 13

42 + 3

7 = 1

5 State whether the function is a probability function.

Yes, this is a probability function since both requirements have been met.

b 1 Substitute each of the x-values into the equation and obtain the corresponding probability.

b When x = 2, p(x) = 16100

= 425

2 Simplify where possible. When x = 3, p(x) = 27100

When x = 4, p(x) = 32100

= 825

When x = 5, p(x) = 25100

= 14

3 Check whether each of the probabilities lie within the restricted interval 0 ≤ Pr(X = x) ≤ 1.

All probabilities lie between 0 and 1 inclusive.

4 Check whether the probabilities sum to 1. 4

25+ 27

100 + 8

25 + 1

4 = 1

5 State whether the function is a probability function.

Yes, this is a probability function since both requirements have been met.

Worked examPle 17

Three balls are selected from a box containing 6 blue balls and 4 yellow balls. If the ball chosen after each selection is replaced before the next selection, fi nd:a the probability distribution for the number of blue balls drawn: i 0 blue balls ii 1 blue ball iii 2 blue balls iv 3 blue ballsb the probability that 3 blue balls are chosen, given that at least one ball was blue.

Think WriTe/draW

a i 1 Draw a tree diagram and list all the possible outcomes with their respective probabilities.

a i

B

Y

B

YYB

YB

B

YYB

Y

B BBB

Outcomes Probability

BBY

BYBBYY

YBBYBY

YYBYYY

4—10

6—10

6—10

6—10 6—

106—10

6—10

6—10

6—10

4—10

6—10

4—10

6—10

6—10

4—10

4—10

4—10

6—10

6—10

4—10

6—10

4—10

4—10

4—10

6—10

4—10

4—10

4—10

6—10

6—10

6—10

6—10

4—10

4—10

4—10

4—10

4—10

4—10

216–––1000

× × =144–––1000

× × =144–––1000

× × =

––1000

–––1000

–––1000

× × =144–––1000

96

96

96

× × =

× × =

× × =64–––

1000× × =


2 Obtain the probability required. Pr(0 blue balls) = 64

1000 (or 0.064)

ii 1 List the required probabilities from the tree diagram obtained in part i.Note: Three outcomes correspond to 1 blue ball.

ii Pr(1 blue ball) = Pr(BYY) + Pr(YBY) + Pr(YYB)

2 Evaluate and simplify. = 3 × 96

1000

= 288

1000 (or 0.288)

iii 1 List the required probabilities from the tree diagram obtained in part i.Note: Three outcomes correspond to 2 blue balls.

iii Pr(2 blue balls) = Pr(BBY) + Pr(BYB) + Pr(YBB)

2 Evaluate and simplify. = 3 × 1441000

= 432

1000 (or 0.432)

iv 1 Obtain the probability required. iv Pr(3 blue balls) = 216

1000 (or 0.216)

2 Place all of the information in a table.

x 0 1 2 3

Pr(X = x) 0.064 0.288 0.432 0.216

3 Check that the probabilities sum to 1.

ΣPr(X = x) = 0.064 + 0.288 + 0.432 + 0.216= 1

b 1 Defi ne the rule. b X XX X

XPr( 3 | 1)

Pr( 3 1)

Pr( 1)= > = = ∩ >

>2 Determine each of the probabilities. Pr(X = 3 ∩ X > 1) = Pr(X = 3) = 0.216

Pr(X > 1) = 0.432 + 0.216= 0.648

3 Substitute values into the rule. Pr(X = 3 | X > 1) = 0.2160.648

4 Evaluate and simplify. = 13

exercise 10B discrete random variables 1 We11 Which of the following represent discrete random variables?

a The number of people at a tennis matchb The time taken to read this questionc The length of the left arms of students in your classd The shoe sizes of twenty peoplee The weights of babies at a maternity wardf The number of grains in ten 250-gram packets of riceg The height of jockeys competing in a certain raceh The number of books in Melbourne libraries

2 We12,13 a If X represents the number of heads obtained in two tosses of a coin, draw up a table that displays the values that the discrete random variable can assume and the corresponding probabilities.b Draw a probability distribution graph of the outcomes in part a.

3 A fair coin is tossed three times and a note is taken of the number of tails.a List the possible outcomes.b List the possible values of the random variable X, representing the number of tails obtained in the

three tosses.c Find the probability distribution of X.d Find Pr(X ≤ 2).

diGiTal doCdoc-9222

SpreadsheetProbability distribution


4 Draw graphs for each of the following probability distributions.

a x 1 2 3 4 5

Pr(X = x) 0.05 0.2 0.5 0.2 0.05

b x 5 10 15 20

Pr(X = x) 0.5 0.3 0.15 0.05

c x 2 4 6 8 10

Pr(X = x) 0.1 0.2 0.4 0.2 0.1

d x 1 2 3 4

Pr(X = x) 0.1 0.2 0.3 0.4

5 We14 Which of the following tables represent a discrete probability distribution?

a x 1 3 5 7 9

Pr(X = x) 0.2 0.3 0.2 0.2 0.1

b x 1 2 3 4 5

Pr(X = x) 0.1 0.1 0.1 0.1 0.1

c x 3 6 9 12 15

Pr(X = x) 0.3 0.2 0.4 0.2 −0.1

d x −4 −3 −1 1 2

Pr(X = x) 0.1 0.1 0.4 0.2 0.2

6 We15 Find the value of k for each of the following discrete probability distributions.

a x 1 2 3 4 5

Pr(X = x) 0.3 0.2 0.2 k 0.1

b x 2 4 6 8 10

Pr(X = x) 0.1 0.1 0.1 0.1 k

c x 0 1 2 3 4

Pr(X = x) k 2k 3k 4k k

d x −2 −1 0 1 2

Pr(X = x) k 0.2 3k 0.3 0.1

7 Find the value of k for the following discrete probability distribution.

a x 1 2 3 4 5

Pr(X = x) k3

13k

13

2 k5 4

13

− k

13

2 7

13

b Explain why one of the values of k had to be discarded.

8 Two fair dice are rolled simultaneously, and X, the sum of the two numbers appearing uppermost, is recorded.a Draw up a table that displays the probability distribution of X, and fi nd:b Pr(X > 9) c Pr(X < 6)d Pr(4 ≤ X < 6) e Pr(3 ≤ X ≤ 9)f Pr(X < 12) g Pr(6 ≤ X < 10).


9 A spinner is numbered from 1 to 5, with each number being equally likely to come up. If X is the random variable representing the number showing on the spinner, find:a the probability distribution of X b the probability of getting an even numberc Pr(X > 2).

10 A fair die is rolled and X is the square of the number appearing uppermost.a Draw up a table that displays the probability distribution of X, and fi nd:b Pr(X < 30) c Pr(X > 10).

11 A fair die is altered so that the 1 is changed to a 5. If X is the random variable representing the number uppermost on the die, find:a the probability distribution of Xb the probability of a number bigger than 2 appearing uppermostc Pr(X = 5 | X > 2).

12 We16a Show that the function p(x) = 190 (8x + 2), where x = 0, 1, 2, 3, 4 is a probability function.

13 We16b Show that the function p(x) = 1160 x2 (x + 2), where x = 1, 2, 3, 4 is a probability function.

14 We17 Three balls are selected from a box containing 4 red balls and 5 blue balls. If the ball chosen after each selection is replaced before the next selection, find, correct to 4 decimal places:a the probability distribution for the number of red balls drawn: i 0 red balls ii 1 red ball iii 2 red balls iv 3 red ballsb the probability that three reds are chosen, given that at least one ball is red.

15 A circular spinner, divided into five equal sectors numbered 1, 2, 3, 4, 5, is spun twice, and the sum of the numbers the pointer lands on is recorded. The following events are then defined.A = ‘an odd number on the fi rst spin’B = ‘an even number on the second spin’C = ‘the sum of the two numbers is odd’D = ‘the sum of the two numbers is at most 7’a List each of the possible outcomes.b Find: i Pr(A) ii Pr(B) iii Pr(C) iv Pr(D).c Find: i Pr(A | B) ii Pr(B | C) iii Pr(C | D).

16 A biased coin is tossed twice. If the probability of obtaining a head is 35:

a fi nd the probability distribution of the number of heads in 2 tossesb show that the sum of the probabilities is 1.

17 A discrete random variable has the following probability distribution:

x 1 2 3 4 5 6 7

Pr(X = x) 0.2 0.11 0.15 0.09 0.17 0.13 0.15

Find:a Pr(X > 3) b Pr(X ≤ 4) c Pr(3 ≤ X ≤ 6)d Pr(2 < X < 5) e Pr(X < 3 | X < 5) f {x: Pr(X < x) = 0.46}g {x: Pr(X ≥ x) = 0.54}.

18 mC Which one of the following random variables is not discrete?a The price, in cents, of a loaf of bread at the local supermarketB The number of runs scored by a batsman in each innings over a seasonC The weight of a baby as he grows over a one-year periodd The number of houses sold by a real estate agent each month for a yeare The number of newspapers recycled by a family each month.

19 mC What is the value of k which will make this table a probability distribution table?

x 1 2 3 4

Pr(X = x) 2k 3k 4k k

a 0 B 1 C 0.1 d 1

9 e −0.1


20 mC Examine the following probability distribution table.

x 4 9 16 25 36

Pr(X = x) 0.16 0.21 0.35 0.08 0.2

Pr(X ≥ 10) is equal to:a 0.38 B 0.84 C 0.35 d 0.28 e 0.63

21 mC The following table represents a discrete probability distribution for a random variable, Y.

x 4 7 10 13

Pr(X = x) d 4d 5d 2k

The value of d is:a

1

9B 1

10C 1

11d 1

12e 1

13

22 mC A coin is biased so that the probability of obtaining a head is 37. If the coin is tossed 3 times the

probability of obtaining exactly 2 heads is:

a 27

343B 108

343C 144

343d 135

343e 64

343

23 mC Which of the following is a probability function?

a p(x) = 0.1, 0.3, 0.4, 0.2, 0.1, x = 0, 1, 2, 3, 4 B p(x) = 166

(3x + 7), x = 0, 1, 2, 3, 4

C p(x) = 140

(5x − 1), x = 1, 2, 3, 4 d p(x) = 120

x2 (4 − x), x = 1, 2, 3

e p(x) = x20

2

(3x − 1), x = 1, 2, 3

24 If the random variable X represents the number of boys in a four-child family:a write the values that X may takeb assuming that the Pr(boy) = 1

2, fi nd the probability distribution of X.

10C measures of centre of discrete random distributionsThe expected value of a discrete random variable, X, is the average value of X. It is also referred to as the mean of X or the expectation.

The expected value of a discrete random variable, X, is denoted by E(X) or the symbol µ (mu). It is defi ned as the sum of each value of X multiplied by its respective probability; that is,

E(X) = x1Pr(X = x1) + x2Pr(X = x2) + x3Pr(X = x3) + . . . + xnPr(X = xn)

= ∑xall

x Pr(X = x)

Note: The expected value will not always assume a discrete value.

Worked examPle 18

Find the expected value of a random variable that has the following probability distribution.

x 1 2 3 4 5

Pr(X = x) 25

110

310

110

110

Think WriTe

1 Write the rule for the expected value. E(X) = x X xPr( )xall

∑ =

2 Substitute the values into the rule. E(X) = 1 × 25 + 2 × 1

10 + 3 × 3

10 + 4 × 1

10 + 5 × 1

10

3 Evaluate. = 2

5 + 2

10 + 9

10 + 4

10 + 5

10

= 22

5

diGiTal doCdoc-9223WorkSHEET 10.1

Concept summary


Units: 3 & 4

AOS: 4

Topic: 1

Concept: 5


Worked examPle 19

Find the unknown probability, a, and hence determine the expected value of a random variable that has the following probability distribution.

x 2 4 6 8 10

Pr(X = x) 0.2 0.4 a 0.1 0.1

Think WriTe

1 Determine the unknown value of a using the knowledge that the sum of the probabilities must total 1.

0.2 + 0.4 + a + 0.1 + 0.1 = 10.8 + a = 1

a = 1 − 0.8= 0.2


∑ =

3 Substitute the values into the rule. E(X) = 2 × 0.2 + 4 × 0.4 + 6 × 0.2 + 8 × 0.1 + 10 × 0.1

4 Evaluate. = 0.4 + 1.6 + 1.2 + 0.8 + 1= 5

Worked examPle 20

Find the values of a and b of the following probability distribution if E(X) = 4.29.

x 1 2 3 4 5 6 7

Pr(X = x) 0.1 0.1 a 0.3 0.2 b 0.2

Think WriTe

1 Write an equation for the values of a and b using the knowledge that the sum of the probabilities must total 1. Call this equation [1].

0.1 + 0.1 + a + 0.3 + 0.2 + b + 0.2 = 10.9 + a + b = 1

a + b = 1 − 0.9a + b = 0.1 [1]


∑ =

3 Substitute the values into the rule. 4.29 = 1 × 0.1 + 2 × 0.1 + 3 × a + 4 × 0.3 + 5 × 0.2 + 6 × b + 7 × 0.2

4 Evaluate and call this equation [2]. = 0.1 + 0.2 + 3a + 1.2 + 1 + 6b + 1.44.29 − 3.9 = 3a + 6b

3a + 6b = 0.39 [2]

5 Solve the equations simultaneously.Multiply equation [1] by 3 and call it equation [3]. Subtract equation [3] from equation [2]. Solve for b. Substitute b = 0.03 into equation [1]. Solve for a.

a + b = 0.1 [1]3a + 6b = 0.39 [2]

3 × (a + b = 0.1)3a + 3b = 0.3 [3]

[2] − [3]: 3b = 0.09b = 0.03

a + 0.03 = 0.1a = 0.1 − 0.03

= 0.07

6 Answer the question. a = 0.07 and b = 0.03


Worked examPle 21

Niki and Melanie devise a gambling game based on tossing three coins simultaneously. If three heads or three tails are obtained, the player wins $20. Otherwise the player loses $5. In order to make a profi t they charge each person $2 to play.a What is the expected gain to the player?b Do Niki and Melanie make a profi t?c Is this a fair game?

Think WriTe/draW

a 1 List the possible outcomes and place all of the information in a table.Note: The ‘Gain’ is from the player’s point of view.

a S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

x 0 1 2 3

Pr(X = x)18

38

38

18

Gain ($) 20 −5 −5 20

2 Write the rule for the expected value.

E(X) = x X xPr( )xall

∑ =

3 Substitute the values into the rule. = 20 × 18 + −5 × 3

8 + −5 × 3

8 + 20 × 1

8

4 Evaluate. = 208

− 158

− 158

+ 208

= 108

= $1.25

5 Answer the question. The player’s expected gain per game is $1.25; however, as each game incurs a cost of $2, the player in fact loses 75c per game.

b Answer the question using the results from a.

b The girls make a profi t of 75c per game.

c Answer the question using the results from a.Note: In a fair game E(X) = 0.

c No, this is not a fair game, since the cost to play each game does not equal the expected gain of each game.

It is important to understand that the expected value signifi es the average outcome of an experiment

and can be used to determine the feasibility of a situation. The previous worked example illustrates that, in the long run, the player will lose on average 75 cents per game; it does not mean the player will lose 75 cents each time the game is played.

expectation theorems

Worked examPle 22

A random variable has the following probability distribution.

x 1 2 3 4

Pr(X = x) 0.25 0.26 0.14 0.35

Find: a E(X) b E(3X) c E(2X − 4) d E(X2).

Think WriTe

a 1 Write the rule for the expected value. a E(X) = x X xPr( )xall

∑ =

2 Substitute the values into the rule. E(X) = 1 × 0.25 + 2 × 0.26 + 3 × 0.14 + 4 × 0.35



3 Evaluate. = 0.25 + 0.52 + 0.42 + 1.4= 2.59

b 1 Write the rule for the expected value. b E(3X) = x X x3 Pr( )xall

∑ =

2 Substitute the values into the rule. E(3X) = (3 × 1) × 0.25 + (3 × 2) × 0.26 + (3 × 3) × 0.14 + (3 × 4) × 0.35

3 Evaluate.Note: 1. The probability remains the same. 2. Each x-value is multiplied by 3

because of the new function, 3x.

= 3 × 0.25 + 6 × 0.26 + 9 × 0.14 + 12 × 0.35

= 0.75 + 1.56 + 1.26 + 4.2 = 7.77

c 1 Write the rule for the expected value. c E(2X − 4) = x X x(2 4) Pr( )xall

∑ − =

2 Substitute the values into the rule. = (2 × 1 − 4) × 0.25 + (2 × 2 − 4) × 0.26 + (2 × 3 − 4) × 0.14 + (2 × 4 − 4) × 0.35

3 Evaluate.Note: 1. The probability remains the same. 2. Each x-value is multiplied by 2 and

then 4 is subtracted from the result, because of the new function, 2x − 4.

= −2 × 0.25 + 0 × 0.26 + 2 × 0.14 + 4 × 0.35

= −0.5 + 0 + 0.28 + 1.4= 1.18

d 1 Write the rule for the expected value. d E(X 2) = x X xPr( )x

2

all∑ =

2 Substitute the values into the rule. = (12) × 0.25 + (22) × 0.26 + (32) × 0.14 + (42) × 0.35

3 Evaluate.Note: 1. The probability remains the same. 2. Each x-value is squared because of

the new function, x2.

= 1 × 0.25 + 4 × 0.26 + 9 × 0.14 + 16 × 0.35

= 0.25 + 1.04 + 1.26 + 5.6= 8.15

The above worked example displays some important points that shall be investigated.For this example, E(X) = 2.59from part b E(3X) = 7.77note that 3E(X) = 3 × 2.59

= 7.77from part c E(2X − 4) = 1.18note that 2E(X) − 4 = 2 × 2.59 − 4

= 1.18.Hence if X is a random variable and a is a constant, its expected value is defi ned by E(aX) = aE(X). Furthermore, if X is a random variable where a and b are constants, then the expected value of a linear function in the form f (X) = aX + b is defi ned by:

E(aX + b) = aE(X) + bIf a = 0 then E(aX + b) = aE(X) + bbecomes E(0X + b) = 0E(X) + b

= b.

These rules are called expectation theorems and are summarised below.

E(aX) = aE(X) where X is a random variable and a is a constant.E(aX + b) = aE(X) + b where X is a random variable and a and b are constants.E(b) = b where b is a constant.E(X + Y) = E(X) + E(Y) where X and Y are both random variables.


These theorems make it easier to calculate the expected values.Finally, from part (d) of the above example it can be seen that:

E(X 2) ≠ [E(X)] 2.

Worked examPle 23

Casey decides to apply for a job selling mobile phones. She receives a base salary of $200 per month and $15 for every mobile phone sold. The following table shows the probability of a particular number of mobile phones, x, being sold per month. What would be the expected salary Casey would receive each month?

x 50 100 150 200 250

Pr(X = x) 0.48 0.32 0.1 0.06 0.04

Think WriTe

Method 11 Defi ne a random variable. Let X = the number of mobile phones sold by Casey in

a month.

2 Write the rule for the expected salary. E(15X + 200) = x X x(15 200)Pr( )xall

∑ + =

3 Substitute the values into the rule. = (15 × 50 + 200) × 0.48 + (15 × 100 + 200) × 0.32 + (15 × 150 + 200) × 0.1 + (15 × 200 + 200) × 0.06 + (15 × 250 + 200) × 0.04

4 Evaluate. = 950 × 0.48 + 1700 × 0.32 + 2450 × 0.1 + 3200 × 0.06 + 3950 × 0.04

= 456 + 544 + 245 + 192 + 158= 1595

5 Answer the question. The expected salary Casey would receive each month would be $1595.

Method 2Using the expectation theorem:

1 Write the rule for the expected salary. E(X) = x X xPr ( )xall

∑ =

2 Substitute the values into the rule. = 50 × 0.48 + 100 × 0.32 + 150 × 0.1 + 200 × 0.06 + 250 × 0.04

3 Evaluate. = 24 + 32 + 15 + 12 + 10= 93

4 Using the fact that E(aX + b) = aE(X) + b fi nd E(15X + 200).

E(15X + 200) = 15E(X) + 200= 15 × 93 + 200= 1595

Note: Using the expectation theorem is quicker because it is easier to evaluate aE(X) + b than E(aX + b).

median and modeThe median is the middle value of the distribution. It is the value such that 50% of the distribution lies to the left of this value, and 50% of the distribution lies to the right.For a random variable, X, the mode is the most commonly occurring value, that is, it is the variable with the highest probability.

Concept summary


Units: 3 & 4

AOS: 4

Topic: 1

Concept: 6


Worked examPle 24

For the following probability distributions, calculate: i the medianii the mode.

a x 3 5 6 8

Pr(X = x) 0.25 0.4 0.15 0.2

b x 5 8 11 14

Pr(X = x) 0.18 0.32 0.2 0.3

Think WriTe

a i 1 To calculate the median, work from the left of the data. Add up the probabilities until the total is 0.5 or greater.

a i Pr(X ≤ 3) = 0.25, which is ≤ 0.5Pr(X ≤ 5) = 0.25 + 0.4

= 0.65, which is ≥ 0.5

2 Write the answer. The median value is 5.

ii 1 To calculate the mode, identify the highest probability in the table.

ii The highest probability is 0.4.That is, Pr(X = 5) = 0.4.

2 Write the answer. The mode is 5.

b i 1 To calculate the median, work from the left of the data. Add up the probabilities until the total is 0.5 or greater.

b i Pr(X ≤ 5) = 0.18, which is ≤ 0.5.Pr(X ≤ 8) = 0.18 + 0.32

= 0.5Similarly Pr(X ≥ 11) = 0.2 + 0.3

= 0.5

2 Calculate the median by calculating the mean of 8 and 11.

Median = +8 112

= 9.5

3 Write the answer. The median is 9.5.

ii 1 To calculate the mode, identify the highest probability in the table.

ii The highest probability is 0.32.That is, Pr(X = 8) = 0.32.

2 Write the answer. The mode is 8.

exercise 10C measures of centre of discrete random distributions 1 We18 Find the expected value of a random variable that has the following probability distribution.

x 0 3 6 9 12

Pr(X = x) 0.21 0.08 0.19 0.17 0.35

2 Find the expected value of a random variable that has the following probability distribution.

x −2 −1 0 1 2 3 4

Pr(X = x) 118

13

118

29

16

118

19

3 We19 Find the unknown probability, a, and hence determine the expected value of a random variable that has the following probability distribution.

x 1 3 5 7 9 11

Pr(X = x) 0.11 0.3 0.15 0.25 a 0.1


4 Find the unknown probability, a, and hence determine the expected value of a random variable that has the following probability distribution.

x −2 1 4 7 10 13

Pr(X = x) 518 a 1

9518

118

29

5 Find the unknown probability, b, and hence determine the expected value of a random variable that has the following probability distribution.

x 0 1 2 3 4 5

Pr(X = x) b 0.2 0.02 3b 0.1 0.08

6 Find the value of k, and hence determine the expected value of a random variable that has the following probability distribution.

x 4 8 12 16 20

Pr(X = x) 6k 2k k 3k 8k

7 If X represents the outcome of a fair die being rolled, find:a the probability distribution of each outcomeb E(X).

8 Two fair dice are rolled simultaneously. If X represents the sum of the two numbers appearing uppermost, find:a the probability distribution of each outcomeb E(X).

9 A fair coin is tossed 4 times. If X represents the number of tails obtained, find:a the probability distribution of each outcomeb E(X).

10 We20 Find the values of a and b of the following distribution if E(X) = 1.91.

x 0 1 2 3 4 5 6

Pr(X = x) 0.2 0.32 a 0.18 b 0.05 0.05

11 Find the values of a and b of the following distribution if E(X) = 2.41.

x 0 1 2 3 4 5

Pr(X = x) 0.2 a 0.23 0.15 b 0.12

12 We21 Lucas contemplates playing a new game which involves tossing three coins simultaneously. He will receive $15 if he obtains 3 heads, $10 if he obtains 2 heads and $5 if he obtains 1 head. However, if he obtains no heads he must pay $30. He must also pay $5 for each game he plays.a What is Lucas’s expected gain?b Should he play the game? Why?c Is this a fair game? Why?

13 Angie plays a game based on tossing three coins simultaneously. She will receive $10 if she obtains 3 tails, $5 if she obtains 2 tails and $5 if she obtains 1 tail. However, if she obtains no tails she must pay $40.a What is Angie’s expected gain?b Should she play the game? Why?c Is this a fair game? Why?

14 X is a discrete random variable with the following probability distribution.

x 2 4 7 k

Pr(X = x) 0.3 0.2 0.4 0.1

Find the value of k if the mean is 5.3.


x −2 3 8 10 14 k

Pr(X = x) 0.1 0.08 0.07 0.27 0.16 0.32

Find the value of k if the mean is 10.98.


16 A coin is biased such that the probability of obtaining a tail is 0.6. If X represents the number of tails in three tosses of the coin, find:a the probability distribution of Xb E(X)c the mode.

17 We22 A random variable has the following probability distribution.

x 1 2 3 4

Pr(X = x) 215

715

13

115

Find:a E(X) b E(4X) c E(2X + 1) d E(X2).

18 A random variable has the following probability distribution.

x 1 2 3 4

Pr(X = x) 0.33 0.25 0.27 0.15

Find:a E(X) b E(4X − 6) c E(X 2 + 1) d E(3X 2).

19 We23 Christian decides to apply for a job selling mobile phones. He receives a base salary of $180 per month and $12 for every mobile phone sold. The following table shows the probability of a particular number of mobile phones, x, being sold per month. What would be the expected salary Christian would receive each month?

x 50 100 150 200 250

Pr(X = x) 0.32 0.38 0.2 0.06 0.04

20 We24 For the following probability distributions, calculate:i the median ii the mode.

a x 1 2 3 4 5

Pr(X = x) 0.25 0.15 0.1 0.1 0.4

b x 4 8 9 13 17

Pr(X = x) 0.06 0.36 0.17 0.29 0.12

c x −2 −1 0 1 2 3 4

Pr(X = x) 1

4

1

16

3

16

1

8

1

8

1

16

3

16

10d measures of variability of discrete random distributionsVarianceVariance is an important feature of probability distributions as it provides information about the spread of the distribution with respect to the mean. If the variance is large, it implies that the possible values are spread (or deviate) quite a distance from the mean. A small variance implies that the possible values are close to the mean. Variance is also called a measure of spread or dispersion.

The variance is written as Var(X) and denoted by the symbol σ 2 (sigma squared). It is defi ned as the expected value (or average) of the squares of the spreads (deviations) from the mean.

The rule for variance is given by: Var(X) = E(X − µ)2

= ∑(X − µ)2 Pr(X = x).

diGiTal doCdoc-9268

SkillSHEET 10.1expected value of a

function of a random variable

diGiTal doCdoc-9224

WorkSHEET 10.2

Concept summary


Units: 3 & 4

AOS: 4

Topic: 1

Concept: 7


Although this rule clearly demonstrates how to obtain the variance, performing the calculation is quite a lengthy process. Hence an alternative rule is used for calculating the variance: Var(X) = E(X − µ)2

= E(X2 − 2µX + µ2)= E(X2) − E(2µX) + E(µ2)= E(X2) − 2µE(X) + E(µ2)= E(X2) − 2µ2 + µ2 since E(X) = µ (the mean)= E(X2) − µ2

= E(X2) − [E(X)]2.

Worked examPle 25

Find the expected value and variance of the following probability distribution table.

x 1 2 3 4 5

Pr(X = x) 0.15 0.12 0.24 0.37 0.12

Think WriTe


∑ =

2 Substitute the values into the rule. = 1 × 0.15 + 2 × 0.12 + 3 × 0.24 + 4 × 0.37 + 5 × 0.12

3 Evaluate. = 0.15 + 0.24 + 0.72 + 1.48 + 0.6= 3.19

4 Calculate E(X 2). E(X 2) = x X xPr( )x

2

all∑ =

= (12) × 0.15 + (22) × 0.12 + (32) × 0.24 + (42) × 0.37 + (52) × 0.12

= 1 × 0.15 + 4 × 0.12 + 9 × 0.24 + 16 × 0.37 + 25 × 0.12

= 0.15 + 0.48 + 2.16 + 5.92 + 3= 11.71

5 Calculate [E(X)]2. [E(X)]2 = 3.192

= 10.1761

5 Calculate Var(X) using the rule for variance.

Var(X) = E(X2) − [E(X)]2

= 11.71 − 10.1761= 1.5339

Worked examPle 26

Find the variance of 2Y + 1 for the following probability distribution table.

y 0 1 2 3

Pr(Y = y) 0.25 0.35 0.2 0.2

Think WriTe

1 Write the rule for the expected value. E(2Y + 1) = y(2 1)xall

∑ + Y yPr( )=

2 Substitute the values into the rule. = (2 × 0 + 1) × 0.25 + (2 × 1 + 1) × 0.35 + (2 × 2 + 1) × 0.2 + (2 × 3 + 1) × 0.2

3 Evaluate. = 1 × 0.25 + 3 × 0.35 + 5 × 0.2 + 7 × 0.2= 0.25 + 1.05 + 1 + 1.4= 3.7

inTeraCTiViTYint-0255measures of variability of discrete random distributions



4 Calculate [E(2Y + 1)]2. [E(2Y + 1)]2 = 3.72

= 13.69

5 Calculate E(2Y + 1)2. E(2Y + 1)2 = 12 × 0.25 + 32 × 0.35 + 52 × 0.2 + 72 × 0.2= 0.25 + 3.15 + 5 + 9.8= 18.2

5 Calculate Var(2Y + 1) using the rule. Var(2Y + 1) = E(2Y + 1)2 − [E(2Y + 1)]2

= 18.2 − 13.69= 4.51

The variance of a linear function can also be calculated by the following rule:

Var (aX + b) = a2Var (X).

For Worked example 26, given that Var(Y) = 1.1275, Var(2Y + 1) can be determined using the above rule: Var(2Y + 1) = 22Var(Y) = 4 × 1.1275 = 4.51 as before.

Worked examPle 27

X is a discrete random variable with the following probability distribution.

x 3 4 6 k

Pr(X = x) 0.15 0.3 0.45 0.1

Find the value of k, a positive integer, if the variance is 1.7475.

Think WriTe


∑ =

2 Substitute the values into the rule. = 3 × 0.15 + 4 × 0.3 + 6 × 0.45 + k × 0.1

3 Evaluate. = 0.45 + 1.2 + 2.7 + 0.1k= 4.35 + 0.1k

4 Calculate E(X2). E(X2) = xx

2

all∑ X xPr( )=

= (32) × 0.15 + (42) × 0.3 + (62) × 0.45 + (k2) × 0.1= 9 × 0.15 + 16 × 0.3 + 36 × 0.45 + k2 × 0.1= 1.35 + 4.8 + 16.2 + 0.1k2

= 22.35 + 0.1k2

5 Calculate [E(X)]2. [E(X)]2 = (4.35 + 0.1k)2

= 0.01k2 + 0.87k + 18.9225

6 Calculate Var(X) using the rule and equate it to the given value of the variance; that is, σ 2 = 1.7475.

Var(X) = E(X2) − [E(X)]2

1.7475 = 22.35 + 0.1k2 − (0.01k2 + 0.87k + 18.9225)1.7475 = 0.09k2 − 0.87k + 3.4275

7 Solve for k. 0.09k2 − 0.87k + 3.4275 − 1.7475 = 00.09k2 − 0.87k + 1.68 = 0

(9k − 24)(k − 7) = 0

k = 24

9 (or 22

3) or k = 7

8 Answer the question. k = 7. Reject the other value of k since the variable is discrete.


Standard deviationAnother important measure of spread is the standard deviation. It is written as SD(X) or denoted by the symbol σ (sigma). The standard deviation is the positive square root of the variance. It is defi ned by the rule:

SD(X) = XVar( )

= 2σ = σ.

Variation and standard deviation are used extensively in many real-life applications involving statistics.Analysis of data would be useless without any information about the spread of the data.

Worked examPle 28

A random variable has the following probability distribution.

x 0 1 2 3

Pr(X = x)1

4

3

81

81

4

Calculate the expected value, the variance and the standard deviation.

Think WriTe

1 Calculate the expected value. E(X) = 0 × 14 + 1 × 3

8 + 2 × 1

8 + 3 × 1

4

= 0 + 38 + 2

8 + 3

4

= 138

2 Calculate [E(X)]2. [E(X)]2 = ( )138

2

= 15764

(≈1.890 625)

3 Calculate E(X 2). E(X2) = 02 × 14 + 12 × 3

8 + 22 × 1

8 + 32 ×

14

= 0 + 38 + 4

8 + 9

4

= 318

4 Calculate Var(X). Var(X) = E(X2) − [E(X)]2

= 318 − 157

64

= 11564

(≈ 1.234 375)

5 Calculate the standard deviation. SD(X) = 1.234375

6 Round the answer to 4 decimal places. = 1.1110

Worked examPle 29

In order to encourage car pooling, a new toll is to be introduced on the Eastgate Bridge. If the car has no passengers, a toll of $2 applies. Cars with one passenger pay a $1.50 toll, cars with two passengers pay a $1 toll and cars with 3 or more passengers pay no toll. Long-term statistics show that the number of passengers (X) follows the probability distribution given below.

x (no. of passengers) 0 1 2 ≥3

Pr(X = x) 0.4 0.35 0.2 0.05

a Construct a probability distribution of the toll paid.b Find the mean toll paid per car.c Find the standard deviation of tolls paid.


Think WriTe

a Construct a table of values of toll information. Probabilities remain the same.

a Let Y = toll to be paid.

y 2 1.5 1 0

Pr(Y = y) 0.4 0.35 0.2 0.05

b 1 Write the rule for the expected value. b E(Y) = ∑ yyall

=Y yPr( )

2 Substitute the values into the rule. = 2 × 0.4 + 1.5 × 0.35 + 1 × 0.2 + 0 × 0.05

3 Evaluate. = 0.8 + 0.525 + 0.2 + 0= 1.525


5 Answer the question. The mean toll is $1.53.

c 1 Calculate E(Y2). c E(Y2) = ∑ yy

2

all

Pr(Y = y)

= (22) × 0.4 + (1.52) × 0.35 + (12) × 0.2 + (02) × 0.05

= 4 × 0.4 + 2.25 × 0.35 + 1 × 0.2 + 0= 1.6 + 0.7875 + 0.2= 2.5875

2 Calculate [E(Y)]2. [E(Y)]2 = 1.5252

= 2.325 625

3 Calculate Var(Y). Var(Y) = E(Y2) − [E(Y)]2

= 2.5875 − 2.325 625= 0.261 875

4 Calculate the standard deviation. SD(Y) = 0.261 875


6 Answer the question. The standard deviation of tolls paid is $0.51.

interpreting the standard deviationA characteristic of many distributions is that approximately 95% of the spread or distribution lies between 2 standard deviations of the mean; that is,

Pr(µ − 2σ ≤ X ≤ µ + 2σ) ≈ 0.95.

It is important to note that when calculating the Pr(µ − 2σ ≤ X ≤ µ + 2σ) for a specifi c distribution an exact value of 0.95 will not always be achieved, but should be close to it.

For many random variables, approximately 95% of the spread of the population lies between 2 standard deviations of the mean, that is,

Pr(µ − 2σ ≤ X ≤ µ + 2σ) ≈ 0.95.

Worked examPle 30

A probability distribution is shown below.Given that µ = 3.38 and σ = 0.946, calculate Pr(µ − 2σ ≤ X ≤ µ + 2σ).

x 2 3 4 5 6

Pr(X = x) 0.14 0.5 0.23 0.1 0.03

Concept summary


Units: 3 & 4

AOS: 4

Topic: 1

Concept: 8


Think WriTe

1 Calculate µ − 2σ. µ − 2σ = 3.38 − 2 × 0.946= 3.38 − 1.892 = 1.488

2 Calculate µ + 2σ. µ + 2σ = 3.38 + 2 × 0.946= 3.38 + 1.892 = 5.272

3 Substitute the values obtained in steps 1 and 2 into the given interval. Since X represents discrete values, Pr(1.488 ≤ X ≤ 5.272) becomes Pr(2 ≤ X ≤ 5).

Pr(µ − 2σ ≤ X ≤ µ + 2σ) = Pr(1.488 ≤ X ≤ 5.272)= Pr(2 ≤ X ≤ 5)= 0.14 + 0.5 + 0.23 + 0.1= 0.97

Note: In this example, 97% of the distribution lies within 2 standard deviations of the mean, which is close to the estimated 95%.

Worked examPle 31

The table below represents the probability distribution of the number of accidents per week in a factory.

x 1 2 3 4 5 6 7 8 9

Pr(X = x) 0.02 0.22 0.18 0.16 0.14 0.07 0.13 0.03 0.05

Given that µ = 4.36 and σ = 2.105 fi nd Pr(µ − 2σ ≤ X ≤ µ + 2σ).

Think WriTe

1 Calculate µ − 2σ. µ − 2σ = 4.36 − 2 × 2.105= 4.36 − 4.21= 0.15

2 Calculate µ + 2σ. µ + 2σ = 4.36 + 2 × 2.105= 4.36 + 4.21= 8.57

3 Substitute the values obtained in steps 1 and 2 into the given interval.Since X represents discrete values, Pr(0.15 ≤ X ≤ 8.57) becomes Pr(1 ≤ X ≤ 8).Note: In this example, Pr(1 ≤ X ≤ 8) = 1 − Pr(X = 9).

Pr(µ − 2σ ≤ X ≤ µ + 2σ) = Pr(0.15 ≤ X ≤ 8.57)= Pr(1 ≤ X ≤ 8)= Pr(X = 1) + Pr(X = 2)

+ Pr(X = 3) + Pr(X = 4)+ Pr(X = 5) + Pr(X = 6)+ Pr(X = 7) + Pr(X = 8)

= 1 − Pr(X = 9)= 1 − 0.05= 0.95

Note: The answer is the estimated one of 95%. In this case, 95% of the distribution lies within 2 standard deviations of the mean.

Worked examPle 32

The probability distribution of X is given by the formula:

Pr(X = x) = x54

2 where x = 2, 3, 4, 5.

Find:a the probability distribution of X as a tableb the expected value of X, correct to 4 decimal placesc the standard deviation of X, correct to 4 decimal placesd Pr(µ − 2σ ≤ X ≤ µ + 2σ), correct to 3 decimal places.



Think WriTe

a 1 Substitute each of the x-values into the equation and obtain the corresponding probability.

a When x = 2, p(x) = 454

= 227

When x = 3, p(x) = 954

= 16

When x = 4, p(x) = 1654

= 827

When x = 5, p(x) = 2554

2 Enter the information into a table. x 2 3 4 5

Pr(X = x)227

16

827

2554

b 1 Calculate the expected value. b E(X) = 2 × 227

+ 3 × 16 + 4 × 8

27 + 5 × 25

54

= 427

+ 36 + 32

27 + 125

54

= 4427


c 1 Calculate [E(X)]2. c [E(X)]2 = (4 427

)2

= 17151729

(≈ 17.2071)

2 Calculate E(X2). E(X2) = 22 × 227

+ 32 × 16 + 42 × 8

27 + 52 × 25

54

= 827

+ 96 + 128

27 + 625

54

= 1819 (≈ 18.1111)

3 Calculate Var(X). Var(X) = E(X2) − [E(X)]2

= 1819 − 17151

729

= 659729

(≈ 0.9040)

4 Calculate the standard deviation. SD(X) = 0.9040 = 0.9 507 776 039


d 1 Calculate µ − 2σ. d µ − 2σ = 4.1481 − 2 × 0.9508 = 4.1481 − 1.9016 = 2.2465

2 Calculate µ + 2σ. µ + 2σ = 4.1481 + 2 × 0.9508 = 4.1481 + 1.9016 = 6.0497

3 Substitute the values obtained in steps 1 and 2 into the given interval. Since X represents discrete values, Pr(2.2465 ≤ X ≤ 6.0497) becomes Pr(3 ≤ X ≤ 6).

Pr(µ − 2σ ≤ X ≤ µ + 2σ) = Pr(2.2465 ≤ X ≤ 6.0497) = Pr(3 ≤ X ≤ 6) = 1 − Pr(X = 2) = 1 − 2

27

= 2527


Note: In this example, 92.6% of the distribution lies within 2 standard deviations of the mean, which is close to the estimated value of 95%.


exercise 10d measures of variability of discrete random distributions 1 We25 Find the expected value and variance of the following probability distribution table.

x 1 2 3 4

Pr(X = x) 0.2 0.4 0.3 0.1


x 2 4 6 8

Pr(X = x) 18

316

916

18

Find:a the expected value, E(X)b the variance of X, Var(X).

3 The cost of a loaf of bread is known to vary on any day according to the following probability distribution.

x $1.20 $1.25 $1.30 $1.35 $1.60

Pr(Y = y) 0.05 0.2 0.1 0.25 0.4

Find:a the expected cost of a loaf of breadb the variance of the cost.

4 We26 Find the variance of 2Y − 1 for the following probability distribution table.

x 0 1 2 3

Pr(Y = y) 0.3 0.2 0.3 0.2


x 2 4 6 8

Pr(X = x) 0.15 0.3 0.42 0.13

Find:a Var(X) b Var(2X) c Var(3X + 1) d Var(−5X + 7).


x 0 1 3 5 7

Pr(X = x) 0.27 0.15 0.13 0.1 0.35

Find:a Var(X) b Var(3X) c Var(10X − 5) d Var(5X − 2).

7 We27 Let X be a discrete random variable with the following probability distribution.

x 2 4 6 k

Pr(X = x) 0.3 0.1 0.5 0.1


8 Let X be a discrete random variable with the following probability distribution.

x 1 k 7 10

Pr(X = x) 0.1 0.2 0.3 0.4


9 We28 A random variable has the following probability distribution.

x 1 2 3 4

Pr(X = x) 14

13

14

16

Calculate the expected value, the variance and the standard deviation.

diGiTal doCdoc-9222SpreadsheetProbability distribution



x 6 7 10 12Pr(X = x) 0.3 0.3 0.2 0.2

Find:a the expected value, E(X)b the variance of X, Var(X)c the standard deviation of X, SD(X), to 2 decimal places.

11 For a random variable, X, E(X) = 12 and E(X2) = 340. Find the standard deviation of X.

12 For a random variable, X, E(X) = 20 and E(X2) = 529. Find the standard deviation of X, to 2 decimal places.

13 We29 In order to encourage car pooling, a new toll is to be introduced on the International Gateway. If the car has no passengers, a toll of $2 applies. Cars with one passenger pay a $1.50 toll, cars with two passengers pay a $1 toll and cars with 3 or more passengers pay no toll. Long-term statistics show that the number of passengers follows the probability distribution given below.

x 0 1 2 ≥3Pr(X = x) 0.5 0.3 0.15 0.05

a Construct a probability distribution of the toll paid.b Find the mean toll paid per car.c Find the standard deviation of tolls paid.

14 We31 The table below represents the probability distribution of the number of accidents per week in a factory.

x 1 2 3 4 5 6 7 8 9Pr(X = x) 0.03 0.21 0.16 0.18 0.14 0.07 0.15 0.01 0.05

Given that µ = 4.35 and σ = 2.08, fi nd Pr(µ − 2σ ≤ X ≤ µ + 2σ).

15 We32 The probability distribution of X is given by the formula, Pr(X = x) = x

35

2

where x = 1, 3, 5. Find:a the probability distribution of X as a tableb the expected value of Xc the standard deviation of X, to 4 decimal placesd Pr(µ − 2σ ≤ X ≤ µ + 2σ).

16 The probability distribution of X is given by the formula, Pr(X = x) = −x 1

50

2

where x = 2, 3, 4, 5. Find:a the probability distribution of X as a tableb the expected value of Xc the standard deviation of Xd Pr(µ − 2σ ≤ X ≤ µ + 2σ).


x 1 2 3 4Pr(X = x) 0.4 0.2 0.2 k

Find:a the value of the constant kb the most likely value of Xc E(X), the mean of Xd Var(X), the variance of Xe SD(X), the standard deviation of X, to 4 decimal placesf Pr(µ − 2σ ≤ X ≤ µ + 2σ).

18 Calculate the values between which 95% of the distribution would be expected to lie where:a µ = 4, σ = 2 b µ = 10, σ = 3c µ = 35, σ = 7 d µ = 21.6, σ = 5.2e µ = 9.7, σ = 0.7 f µ = 17

12, σ = 2

13.


19 Two fair dice are rolled and the outcomes are noted. If X represents the sum of the two numbers showing, find:a the expected value of Xb the variance of X, to 2 decimal placesc Pr(µ − 2σ ≤ X ≤ µ + 2σ), correct to 2 decimal places.

20 For the spinner shown at right, X represents the number obtained. Find:a the probability distribution of Xb the expected value of Xc the standard deviation of X, to 2 decimal placesd the probability that the number is 4, given that it is not 1.

21 mC The following table represents a discrete probability distribution for a random variable, X.

x 0 1 2 3

Pr(X = x) k 2k 3k 4k

The standard deviation of x is:a 1.0 B 1.2 C 1.8 d 2.0 e 2.2Questions 22 and 23 refer to the following information. The probability distribution of X is given in the table below.

x 0 3 6 9

Pr(X = x) 0.4 0.3 0.1 0.2

22 mC The variance and standard deviation of X, respectively, are:a 3.9, 15.21 B 26.1, 3.3 C 26.1, 3.9 d 26.1, 5.11 e 11.61, 3.41

23 mC Var (6X − 3) is equal to:a 91.26 B 541.56 C 417.96 d 939.6 e 140.4

32

4

21

1

2 3


SummaryProbability revision • Outcomes are results of experiments.

• The set of all possible outcomes of an experiment is called the sample space and is denoted by ε, and each possible outcome is called a sample point.

• A subset of the sample space is known as an event.• The union (symbol ∪) of two events A and B implies a combined event, that is, either event A or

event B or both occurring. Common elements are written only once.• The intersection (symbol ∩) of two events A and B is represented by the common sample points of

the two events.• Venn diagrams involve drawing a rectangle that represents the sample space and a series of circles

that represent subsets of the sample space. They provide a visual representation of the information at hand and clearly display the relationships between sets.

• The probability of an event occurring is defi ned by the rule:

Pr(A) = number of favourable outcomes .


• The probability of an event occurring lies within the restricted interval 0 ≤ Pr(A) ≤ 1.• The individual probabilities of a particular experiment will sum to 1; that is, Σ p(x) = 1.• The addition rule of probability is defi ned by the rule Pr(A ∪ B) = Pr(A) + Pr(B) − Pr(A ∩ B).• If two events A and B are mutually exclusive, then Pr(A ∩ B) = 0 and therefore the addition rule

becomes Pr(A ∪ B) = Pr(A) + Pr(B).• If two events A and B are independent, then Pr(A ∩ B) = Pr(A) × Pr(B).• Karnaugh maps and probability tables summarise all combinations of two events (for example A

and B) and their complements (for example A′ and B′).• Conditional probability is defi ned by the rule Pr(A | B) =

∩A B

B

Pr( )

Pr( ), where Pr(B) ≠ 0. This can be

transposed to Pr(A ∩ B) = Pr(A | B) × Pr(B).• Tree diagrams are useful tools in solving probability tasks as they display each of the possible

outcomes along with their respective probabilities.• A combination is defi ned by nCr , that is, the number of selections of n different objects taken r at a time.

discrete random variables

• A random variable is one whose value is determined by the outcome of an experiment.• Discrete random variables generally deal with number or size and are able to be counted.• Two important characteristics satisfy all discrete probability distributions:

1. Each probability lies in a restricted interval 0 ≤ Pr(X = x) ≤ 1.2. The probabilities of a particular experiment sum to 1; that is,

∑ = =X xPr( ) 1

If these two characteristics are not satisfi ed, then there is no discrete probability distribution.

measures of centre of discrete random distributions

• The expected value of a discrete random variable, X, is denoted by E(X) or the symbol µ (mu). It is defi ned by the rule:

E(X) = ∑ =x X xPr ( )xall

.

• A game is considered fair if the cost to play the game is equal to the expected gain.• A fair game is one in which E(X) = 0.• The expected value of a linear function can be calculated using the expectation theorems:

E(aX) = aE(X)E(aX + b) = aE(X) + bE(b) = bE(X + Y) = E(X) + E(Y).Note: E(X2) ≠ [E(X)]2

• The median is the middle value of a distribution.• The mode is the variable with the highest probability.


measures of variability of discrete random distributions

• The variance is denoted by Var(X) or the symbol σ 2 (sigma squared).• It is defi ned by the rule:

Var(X) = E(X2) − [E(X)]2.

• The variance of a linear function can also be calculated by the following rule:

Var(aX + b) = a2Var(X).

• The standard deviation is written as SD(X) or denoted by the symbol σ.• It is defi ned by the rule:

SD(X) = XVar( )

= σ 2 .

• Approximately 95% of the spread of the population in many distributions lies between 2 standard deviations of the mean, that is,

Pr(µ − 2σ ≤ X ≤ µ + 2σ) ≈ 0.95.


Chapter review 1 The Santaroos have 3 soccer teams, A, B and C, entered in the interschool championships. Each team is

entered in a separate division. The probability of each team winning their particular final is given as follows:

Pr(Team A wins) = 12 Pr(Team B wins) = 1

3 Pr(Team C wins) = 3

4.

Find the probability that:a none of the Santaroos teams winb one of the Santaroos teams winsc two of the Santaroos teams wind each of the Santaroos teams win their particular fi nal.

2 Thirty students were required to complete a logic puzzle. The time taken to complete the puzzle was recorded in the table below.

Time taken (whole number of minutes) 3 4 5 6 7

Number of students 4 8 10 2 6

For the information given:a what proportion of students completed the puzzle in less than 5 minutes?b what proportion of students took more than 5 minutes to complete the puzzle?

3 The probability distribution for the service time at a bakery is given below.

Service time (whole number of minutes) 1 2 3 4 5

Probability 0.2 0.3 0.2 0.2 0.1

a What is the probability that the service time for a customer is: i 2 minutes? ii 2 minutes or less? iii more than 2 minutes? iv not more than 4 minutes, given it is more than 1 minute?b What is the expected value for the service time?c If 50 customers were served at the bakery in a morning, how many of these would you expect to

take 4 minutes to be served?

4 The probability distribution of X is given by the formula, Pr(X = x) = x30

2

, where x = 1, 2, 3, 4. Find:a the probability distribution of X as a tableb the expected value of X.

5 A player rolls a fair die. If the player gets a 1 on the first roll, she rolls again and her score is the sum of the two results, otherwise her score is the result of the first roll. The die cannot be thrown more than twice. Find:a the probability distributionb the expected scorec Pr(X < µ).

6 Tayah and Sandy are playing a game where a biased die is rolled. The probabilities of rolling each number are Pr(1) = 0.2, Pr(2) = Pr(3) = Pr(5) = 0.1, Pr(4) = 0.3 and Pr(6) = 0.2.They have to pay $1.00 to play. If a 2, 3 or 5 is rolled, they win $3. If a 1 or 6 is rolled, they get their money back, and if a 4 is rolled, they do not receive any money. Is this a fair game to play?

7 A game of ‘three-up’ is played where three coins are tossed simultaneously. A player must pay $2 to play the game. If three heads come up, the player collects $6. If two heads come up, the player collects $3. Is it a fair game?

8 Kylie is about to compete in her club’s archery finals. If she is equally likely to hit any point on the board and never misses the target, find:a her expected score from i 1 shot at the target shown ii 5 shots at the target shownb her probability of getting fi ve points on every one of her 5 shots.

9 For the following probability distributions, calculate: i the medianii the mode.

ShorT anSWer

2 points

11 points

5 points

30 cm

10 cm

20 cm


a x 3 4 5 6 7

Pr(X = x) 0.1 0.45 0.09 0.26 0.1

b x 1 4 6 8 11

Pr(X = x) 15

15

110

25

110

10 At Fast Eddy’s Drive-In Theatre the cost is $5 per car, plus $1 per occupant. The variable X represents the number of people in any car and is known to follow the probability distribution below.

x 2 3 4 5

Pr(X = x) 0.4 0.2 0.3 0.1

Find:a the expected cost per carb Fast Eddy’s expected profi t if 100 cars enter and costs for wages, electricity, etc. are $300c the mode.


x 1 3 5 n

Pr(X = x) 0.1 0.25 0.35 0.3

Find the value of n if the mean is 4.7.

12 Five thousand ‘scratch-and-match’ tickets are to be sold for $2 each. The tickets offer the following prizes: 1 prize of $5000 2 prizes of $500 20 prizes of $50 100 prizes of $10.Find:a the expected loss per ticketb the profi t made by organisers, that is, the housec the house percentage.

13 A raffle is to be drawn from 500 tickets. Each ticket was purchased for $1, with first prize being $200, second prize $150 and third prize $100. Find:a the expected loss per ticketb the profi t made by organisers, that is, the housec the house percentage.

14 Pr(X = x) = +x x( 1)

40 for 0 ≤ x ≤ n

a Find the value of n.b Find the probability distribution for X as a table.c Find the expected value of X.

mUlTiPleChoiCe

1 If Pr(A) = 0.65, Pr(B) = 0.37 and Pr (A ∩ B) = 0.28, then Pr(A ∪ B) is equal to:a 0.93 B 0.56 C 0.09 d 0.74 e 1.02

2 If Pr(A) = 0.47, Pr(B) = 0.27 and Pr (A ∩ B) = 0.19, then Pr(A | B) is equal to:a 0.57 B 0.40 C 0.70 d 0.43 e 0.30

3 The probability that Fiona attends an aerobics class is 0.60, and the probability that Kath attends an aerobics class is 0.85. If these two events are independent, the probability of one of these two people attending an aerobics class is:a 0.60 B 0.85 C 0.51 d 0.34 e 0.43

4 Which of the following random variables is discrete?a The number of runs scored by Sir Donald Bradman in his cricketing careerB The weight of people in an elevatorC The life span of a fl yd The volume, in litres, of water in the Yarra Rivere The time, in hours, for a student to complete a Mathematical Methods test


5 Which of the following does not represent a probability distribution?

a x 2 4 6 8 10

Pr(X = x) 0.2 0.3 0.2 0.2 0.1

B x 1 2 3 4 5

Pr(X = x) 0.1 0.24 0.03 0.56 0.07

C x 3 6 9 12 15

Pr(X = x) 0.36 0.12 0.4 0.02 0.1

d x −4 −3 −1 1 1

Pr(X = x) −0.1 0.2 0.4 0.2 0.3

e x −7 −5 −3 −1 1

Pr(X = x) 0.09 0.12 0.41 0.18 0.2


x 0 1 2 3 4 5

Pr(X = x) 0.1 0.2 0.3 0.1 0.2 0.1

The probability that the variable is an odd number, given that it is less than 4 is:

a 12

B 710

C 45

d 37

e 13

7 The value of k for the following probability distribution is:

x 1 2 3 4 5

Pr(X = x) 0.3 5k 0.2 3k 0.1

a 0.15 B 0.05 C 0.25 d 0.4 e 1

8 A die is biased so that Pr(X = 1) = Pr(X = 2) = 0.1 and Pr(X = 3) = Pr(X = 4) = Pr(X = 5) = Pr(X = 6) = 0.2. A game is played where a player rolls the die. The player receives $5 if a number greater than 4 is obtained but must pay $2 if a number less than or equal to 4 comes up. The expected result for the player for each roll is:a a loss of 80c B a loss of $1.40 C a win of 80cd a loss of $1.40 e a win of $1.20

9 Gertrude’s Gambling House offers patrons a card game which uses a deck comprising four aces, three kings, two queens and one jack. A player draws a card at random. If a jack is drawn the player wins $5, while a queen results in a win of $2. However, if the player draws an ace, a loss of $1 is incurred. On average, Gertrude’s Gambling House wins 40c each time the game is played. If a king is drawn, a player must pay:a $1.00 B $3.00 C $1.50 d $2.00 e $2.50

10 Sam’s chance of getting a bullseye while playing darts is 0.1 and his chance of missing the board altogether is 0.2. Sam collects $2 for a bullseye or 20c for hitting any other part of the board. If the game is to be fair, missing the board altogether means Sam must pay:a $1.70 B $2.70 C $2.20 d $1.50 e $1.85

11 The toll for a new freeway is $2 per car and 50c per occupant. Long-term surveys show that X, the number of occupants per car, is distributed as follows.

x 1 2 3 4 5

Pr(X = x) 0.3 0.3 0.2 0.1 0.1

The expected toll for each car is:a $3.20 B $3 C $3.50 d $2.95 e $3.65


Questions 12 and 13 refer to the following probability distribution.

x 1 2 3 4 5

Pr(X = x) 0.3 0.3 0.2 0.1 0.1

12 E(X) is equal to:a 2.5 B 3.1 C 5.2 d 2.4 e 2.6

13 E(5X − 8) is equal to:a 18 B 4.5 C 4 d 7.5 e 5

Questions 14, 15 and 16 refer to the following probability distribution.

x −2 −1 0 1 2

Pr(X = x) 0.1 0.25 0.2 0.15 0.3

14 E(X) is equal to:a 1.2 B 0.5 C 1.4 d 1.5 e 0.3

15 Var(X) is equal to:a 2 B 1.7 C 1.91 d 0.61 e 2.11

16 Var(3X + 2) is equal to:a 8 B 17.19 C 7.73 d 15.3 e 2.7


x 3 6 9 12

Pr(X = x) 0.21 0.35 0.17 0.27

The mean and standard deviation are:a µ = 6 and σ = 3.3 B µ = 7.5 and σ = 10.89C µ = 7.6 and σ = 3.3 d µ = 7.5 and σ = 3.3e µ = 6 and σ = 10.89

18 Let X be a random variable with the following probability distribution.

x 1 2 3 4

Pr(X = x) 0.2 0.2 0.4 0.2

The values between which 95% of the distribution lie for the discrete random variable X are:a [1, 3] B [1, 4] C [2, 4] d [1, 2] e [2, 3]

exTended reSPonSe

1 Bob’s Bakery makes four different types of doughnut, each at a different price, depending on the ingredients used. A school canteen buys all its doughnuts from Bob and is currently estimating budgets for the upcoming financial year. Types of doughnut and their prices are listed below, along with their popularity (expressed as a probability).

Doughnut type Cost per box (2 dozen) Probability

Jam $14.4014

Iced $15.6014

Cinnamon $12.00310

Iced jam $18.0015

Find:a the mean price per boxb the standard deviation per boxc the average price per doughnut.


The canteen wants to make a 14% profi t on costs. Find:d the cost of a doughnut at the canteen, if all doughnuts

are to be sold for the same amounte the average profi t per box.

2 Amina plays roulette, a game where a wheel containing 37 slots numbered 0–36 is spun and the winning number is the one in which a ball lodges when the wheel stops spinning. Amina plays three different games:a First she bets $20 on her favourite number coming up at

Casino-nominated odds of 35:1 against. i How much would Amina collect if her number came up? ii Find her expected win or loss for the game. iii Is this game fair?b In the second game Amina bets $20 on an even number coming up at Casino-nominated odds of

1:1 (even money chance). i How much would Amina collect if an even number came up? ii Find her expected win or loss for the game. iii Is this game fair?c In game number three, Amina bets $20 on a line of 12; that is, if numbers 1–12 come up, she

wins. The Casino-nominated odds for this game are 2:1 against. i How much would Amina collect if one of her numbers came up? ii Find her expected win or loss for the game. iii Is this game fair?d What is the house percentage for these games?

3 A door-to-door telecommunications representative has recorded her day-by-day sales figures over a period of time. She knows that her probability of selling X packages on any one day follows the probability distribution shown in the table.

x 0 1 2 3 4 5 >5

Pr(X = y) 2t2 3t 2t2 2t 4t2 + t t 0

a Find the value of t.b Find the probability that she sells at least 2 packages on any one day.c Find the probability that she sells at most 4 packages on any one day.d Find the number of packages she can expect to sell each day.e Calculate the Var(X) and standard deviation of X, correct to 4 decimal places.f Find Pr(µ − 2σ ≤ X ≤ µ + 2σ).g If the representative receives a commission of $25 per package sold and a bonus of $200 if she

sells 4 or more packages in one day, fi nd her expected daily earnings from commissions and bonuses.

h Given that the representative will sell at least two packages tomorrow, fi nd the probability that she will get her $200 bonus.

0 154

217

613

1181024

33

2031

2229

28

3526 32

19

2125

3427

36

3023516

1

149

187

123

Units: 3 & 4

AOS: 4

Topic: 1

Practice VCE exam questions

Use StudyON to access all exam questions on this topic since 2002.

diGiTal doCdoc-9225

Test YourselfChapter 10


ICT activitiesChapter openerdiGiTal doC

• 10 Quick Questions doc-9221: Warm up with ten quick questions on applications of discrete random variables. (page 465)

10a Probability revisionTUTorialS

• We7 eles-1226: Watch a worked example on calculating conditional probability. (page 473)

• We10 eles-1227: Watch a worked example on considering combinations when conducting probability experiments. (page 476)

10B discrete random variablesTUTorial

• We16 eles-1228: Watch a worked example on verifying probability functions. (page 482)

diGiTal doCS• Spreadsheet doc-9222: Investigate probability distributions.

(page 484)• WorkSHEET 10.1 doc-9223: Determine probabilities given discrete

data and distributions including conditional probability and graphs. (page 487)

10C measures of centre of discrete random distributionsTUTorial

• We22 eles-1229: Watch a worked example on calculating measures of centre. (page 489)

diGiTal doCS• SkillSHEET 10.1 doc-9268: Practise finding the expected value of a

function of a random variable. (page 494)• WorkSHEET 10.2 doc-9224: Calculate probabilities, expected values

and variance for discrete probability distributions. (page 494)

10d measures of variability of discrete random distributionsinTeraCTiViTY

• Measures of variability of discrete random distributions int-0255: Consolidate your understanding of measures of centre and variability. (page 495)

TUTorialS• We26 eles-1230: Watch a worked example on calculating

variance. (page 495)• We32 eles-1231: Watching a worked example on constructing a

probability distribution table. (page 499)

diGiTal doC• Spreadsheet doc-9222: Investigate probability distributions.

(page 501)

Chapter reviewdiGiTal doC

• Test Yourself doc-9225: Take the end-of-chapter test to test your progress. (page 510)

To access eBookPLUS activities, log on to www.jacplus.com.au


Answers CHAPTER 10

diSCreTe random VariaBleS exercise 10a Probability revision

1 a 118 b 1

36 c 16 d 56 e 7

12 f 12 g 5

12

2 a i 14 ii 512 iii 13

b 14 + 512 + 13 = 1

c i 23

ii 23

iii 14

iv 1

3 a i 18

b 38

c 38

d 18

e 12 4 a 3

8 b 1

4 c 3

8 d 5

8 5 0.7 6 0.1 7 a 0.2 b 0.8 8 a 2

5 b 13

40

c 1740

d 320

e 38

9 a 13

b 12 c No d Yes 10 No 11 Yes

12 a Column

1Column

2Column

3B B′

Row 1 A 0.22 0.10 0.32

Row 2 A′ 0.08 0.60 0.68

Row 3 0.30 0.70 1

b i Pr(A ∩ B) represents an unfi t smoker, Pr(A ∩ B) = 0.22.

ii Pr(A′ ∩ B′) represents a fi t non-smoker, Pr(A′ ∩ B′) = 0.60.

c i 0.10 ii 0.08 iii 0.32 iv 0.70

13 a 23

b 12

c 13

d No e Yes 14 a 0.7 b 0.4 c 0.5 15 a 0.105 b 0.39

c 0.495 d 733

16 a 0.7 b 0.3 c 2735

17 a 2564

b 964

c 1532

d 1732

18 C 19 D 20 D

21 B 22 14

23 a 0.4 b 0.5 c 0.74

d 0.26 e 0.76 f 1825

g 35

h 1215

i 720

24 a 25 b 115

c 815

d 715

25 a 2572

b 1966

26 A 27 B 28 a 55

64 b 6

7 29 a 0.3277 b 0.64 30 0.5 31 2

9 32 22

31

33 a 0.1035 (4 d.p.) b 0.3929 (4 d.p.) c 0.2746 (4 d.p.)

exercise 10B discrete random variables 1 a Discrete b Continuous c Continuous d Discrete e Continuous f Discrete g Continuous h Discrete 2 a x 0 1 2

Pr(X = x) 14

12

14

b

x0 1 2

Pr(X = x)3–41–21–4

3 a HHH, HHT, HTH, HTT, THH, THT, TTH, TTT

b x = 0, 1, 2, 3 c x 0 1 2 3

Pr(X = x) 18

38

38

18

d 78

4 a

x

Pr(X = x)

0.10.20.3

0.50.4

b

x

Pr(X = x)

0.20.1

0.3

0.50.4

5 10 2015

c

x

Pr(X = x)

0.1

2 4 6 8 10

0.2

0.30.4

d

x

Pr(X = x)

0.1

1 2 3 4

0.2

0.3

0.4

5 a, d 6 a 0.2 b 0.6 c 1

11 d 0.1

7 a x 1 2 3 4 5

Pr(X = x) 313

113

113

113 7

13

b k = −5 was discarded as it would result in negative probabilities.

8 a

x 2 3 4 5 6 7 8 9 10 11 12

Pr(X = x) 136

118

112

19

536

16

536

19

112

118

136

b 16 c 518 d 7

36

e 2936

f 3536 g 59

9 a x 1 2 3 4 5

Pr(X = x) 15

15

15

15 1

5

b 25 c 35

10 a x 1 4 9 16 25 36

Pr(X = x) 16

16

16

16

16

16

b 56 c 12 11 a x 2 3 4 5 6

Pr(X = x) 16

16

16

13

16

b 56 c 25

12 x 0 1 2 3 4

Pr(X = x) 145

19

15

1345

1745

13 x 1 2 3 4

Pr(X = x) 3160

110

932

35

14 a x 0 1 2 3Pr(X = x) 0.1715 0.4115 0.3292 0.0878

b 0.1060 15 a (1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (1, 2),

(2, 2), (3, 2), (4, 2), (5, 2), (1, 3), (2, 3), (3, 3), (4, 3), (5, 3), (1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (1, 5), (2, 5), (3, 5), (4, 5), (5, 5)

b i Pr(A) = 35 ii Pr(B) = 25

iii Pr(C) = 1225 iv Pr(D) = 1925

c i Pr(A | B) = 35 ii Pr(B | C) = 12 iii Pr(C | D) = 1019 16 a x 0 1 2

Pr(X = x) 425

1225

925

b 425 + 1225 + 925 = 1

17 a 0.54 b 0.55 c 0.54

d 0.24 e 3155 f 4 g 4 18 C 19 C 20 E 21 D 22 B 23 D 24 a x = 0, 1, 2, 3, 4 b x 0 1 2 3 4

Pr(X = x) 116

14

38

14

116

exercise 10C measures of centre of discrete random distributions 1 7.11 2 13

18 3 a = 0.09, E(X) = 5.42 4 a = 118 , E(X) = 51

3

5 b = 0.15, E(X) = 2.39 6 k = 0.05, E(X) = 13 7 a x 1 2 3 4 5 6

Pr(X = x) 16

16

16

16

16

16

b 312


8 a x 2 3 4 5 6 7 8 9 10 11 12

Pr(X = x) 136

118

112

19

536

16

536

19

112

118

136

b 7 9 a x 0 1 2 3 4

Pr(X = x) 116

14

38

14

116

b 2 10 a = 0.15, b = 0.05 11 a = 0.1, b = 0.2 12 a $3.75 b No, because although his expected gain

is $3.75 per game, he must pay $5 to play each game. Therefore his loss per game will be $1.25.

c No, because the expected gain is not equal to the initial cost of the game.

13 a $0 b No, because even though her expected

gain (that is, her average gain per game in the long run) is $0, she could still lose $40 in one single game.

c Yes, because E(X) = 0. 14 11 15 17 16 a

x 0 1 2 3

Pr(X = x) 8125

36125

54125

27125

OR 0.064 0.288 0.432 0.216

b 145 c 2

17 a 213 b 91

3 c 523 d 6 1

15

18 a 2.24 b 2.96 c 7.16 d 18.48 19 $1452 20 a i 3.5 ii 5 b i 9 ii 8 c i 0.5 ii –2

exercise 10d measures of variability of discrete random distributions 1 2.3, 0.81

2 a 538 b 255

64

3 a $1.42 b $0.02 4 4.96 5 a 3.2364 b 12.9456 c 29.1276 d 80.91

6 a 8.7899 b 79.1091 c 878.99 d 219.7475 7 10 8 5

9 213, 1.0556, 1.0274

10 a 8.3 b 5.41 c 2.33

11 14 12 11.36 13 a y 2 1.5 1 0

Pr(Y = y) 0.5 0.3 0.15 0.05

b $1.60 c $0.51 14 0.95 15 a x 1 3 5

Pr(X = x) 135

935

57

b 41325 c 1.0443 d 3435

16 a x 2 3 4 5

Pr(X = x) 350

425

310

1225

b 415 c 21

5 (≈ 0.9165)

d Pr(2.3669 ≤ X ≤ 6.0330) = Pr(3 ≤ X ≤ 6) = 0.94 17 a 0.2 b 1 c 2.2 d 1.36 e 1.1662 f 1 18 a 0 ≤ X ≤ 8 b 4 ≤ X ≤ 16 c 21 ≤ X ≤ 49 d 11.2 ≤ X ≤ 32

e 8.3 ≤ X ≤ 11.1 f 1256 ≤ X ≤ 221

6

19 a 7 b 5.83 c 0.94 20 a x 1 2 3 4

Pr(X = x) 14

38

14

18

b 2.25 c 0.97 d 16 21 A 22 E 23 C

ChaPTer reVieWShorT anSWer

1 a 112 b 9

24 c 512 d 18

2 a 25 b 415

3 a i 0.3 ii 0.5 iii 0.5 iv 7

8(0.875)

b 2.7 min c 10

4 a x 1 2 3 4

Pr(X = x) 130

215

310

815

b 313

5 a x 2 3 4 5 6 7

Pr(X = x) 736

736

736

736

736

136

b 4 112 c 7

12 6 No, it is not a fair game. 7 No 8 a i 4 ii 20

b 1243

9 a i 4 ii 4 b i 7 ii 8

10 a $8.10 b $510 c 2 11 7 12 a 40 cents b $2000 c 20%

13 a 10 cents b $50 c 10% 14 a 4 b x 0 1 2 3 4

Pr(X = x) 0 120

320

310

12

c 314

mUlTiPle ChoiCe

1 D 2 C 3 E 4 A 5 D 6 D 7 B 8 C 9 B 10 A 11 A 12 D 13 C 14 E 15 C 16 B 17 D 18 B

exTended reSPonSe

1 a $14.70 b $2.14 c 61.25 cents d 70 cents e $2.10 2 a i $720 ii Loss of 54 cents iii No b i $40 ii Loss of 54 cents iii No c i $60 ii Loss of 54 cents iii No d 2.7%

3 a 18

b 1932 c 7

8 d 2.5625 packages

e Var(X) = 2.3086, SD(X) = 1.5194 f 1 g $126.56 h 10

19

discrete random variables...chapter 10 • discrete random variables 467 2 list all the possible...

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