chapter 3: discrete probability distributions...

Civil Engineering Department: Engineering Statistics (ECIV 2005)

Engr. Yasser M. Almadhoun

Chapter 3: Discrete Probability Distributions

(Cont’d)

Section 3.3: The Hypergeometric Distribution

Problem (01): Let X have a hypergeometric distribution with N = 11, r = 6, and n = 7.

Calculate:

(a) P(X = 4) (b) P(X = 5) (c) P(X ≤ 3)

(d) E(X) (e) Var(X)

(Problem 3.3.1 in textbook)

Solution:

P(X = 4) =(𝑟𝑥) × (

𝑁 − 𝑟𝑛 − 𝑥

)

(𝑁𝑛)

=(64) × (

11 − 67 − 4

)

(116

)= 0.4545

P(X = 5) =(𝑟𝑥) × (


)

(𝑁𝑛)

=(65) × (

11 − 67 − 5

)

(116

)= 0.1818

P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

=(𝑟𝑥) × (


)

(𝑁𝑛)

=(60) × (

11 − 67 − 0

)

(116

)+

(61) × (

11 − 67 − 1

)

(116

)



+(62) × (

11 − 67 − 2

)

(116

)+

(63) × (

11 − 67 − 3

)

(116

)

= 0.5000

E(X) =nr

N=

7 × 6

11= 3.8182

Var(X) = (𝑁 − 𝑛

𝑁 − 1) (𝑛) (

𝑟

𝑛) (1 −

𝑟

𝑁)

= (11 − 7

11 − 1) (7) (

6

7) (1 −

6

11)

= 1.0910

Problem (02): A committee consists of eight right-wing members and seven left-wing

members. A subcommittee is formed by randomly choosing five of the

committee members. Draw a line graph of the probability mass function of

the number of right-wing members serving on the subcommittee.


Solution:

P(X = x) =(𝑟𝑥) × (


)

(𝑁𝑛)



Problem (03):

A box contains 17 balls of which 10 are red and 7 are blue. A sample of 5

balls is chosen at random and placed in a jar. Calculate the probability that:

(a) The jar contains exactly 3 red balls.

(b) The jar contains exactly 1 red ball.

(c) The jar contains more blue balls than red balls.


Solution:

P(X = x) =(𝑟𝑥) × (


)

(𝑁𝑛)

P(X = 3) =(𝑟𝑥) × (


)

(𝑁𝑛)

=(103

) × (17 − 105 − 3

)

(175

)= 0.4072



P(X = 1) =(𝑟𝑥) × (


)

(𝑁𝑛)

=(101

) × (17 − 105 − 1

)

(175

)= 0.0566

P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)

=(𝑟𝑥) × (


)

(𝑁𝑛)

=(100

) × (17 − 105 − 0

)

(175

)

+(101

) × (17 − 105 − 1

)

(175

)

+(102

) × (17 − 105 − 2

)

(175

)

= 0.3145

Problem (04): A jury of 12 people is selected at random from a group of 16 men and 18

women.

(a) What is the probability that the jury contains exactly 7 women?

(b) Suppose that the jury is selected at random from a group of 1600

men and 1800 women. Use the binomial approximation to the

hypergeometric distribution to calculate the probability that in this

case the jury contains exactly 7 women.


Solution:



P(7 women) =(𝑟𝑥) × (


)

(𝑁𝑛)

=(

1612 − 7

) × (187

)

(16 + 18

12)

+(165

) × (187

)

(3412

)= 0.2535

∼ ∼ ∼

𝑃(𝑋 = 7) = (𝑛𝑥) × 𝑝𝑥 × (1 − 𝑝)𝑛−𝑥

= (127

) × (0.53)7 × (1 − 0.53)12−7

= 0.2131

Problem (05): There are 11 items of a product on a shelf in a retail outlet, and unknown

to the customers, 4 of the items are outdated. Suppose that a customer takes

3 items at random.

(a) What is the probability that none of the outdated products are

selected by the customer?

(b) What is the probability that exactly 2 of the items taken by the

customer are outdated?


Solution:



P(X = 0) =(𝑟𝑥) × (


)

(𝑁𝑛)

=(40) × (

11 − 43 − 0

)

(113

)= 0.2121

P(X = 2) =(𝑟𝑥) × (


)

(𝑁𝑛)

=(42) × (

11 − 43 − 2

)

(113

)= 0.0424

Problem (06): A box contains 100 calculators. 80 of them are made in Britain. While the

rest are made in China. Two calculators are sampled successively at

random from the box without replacement.

(a) Compute the probability that the first one is made in China.

(b) If it was known that first one is made in China, what is the

probability that the second one is also made in China.

(c) Compute the probability that both calculators are made in the same

country.

(d) Seven calculators are selected at random without replacement from

the box. Determine the probability that four of them are made in

Britain.

(e) A sample of eight calculators is to be selected at random without

replacement from the box. How many different samples are there

of size eight that contain exactly two Chinese calculators.

(f) A sample of eight calculators is to be selected at random without

replacement from the box. What is the probability that the chosen

sample contains more Chinese calculators than British calculators.

(Question 2: (9 points) in Midterm Exam 2012)

Solution:



P(1st C) =20

100= 0.2000

P(1st C ∩ 2nd C) =20

100×

19

99= 0.0384

P((1st B ∩ 2nd B) ∪ (1st C ∩ 2nd C))

= (80

100×

79

99) + (

20

100×

19

99) = 0.6768

P(X = 4) =(𝑟𝑥) × (


)

(𝑁𝑛)

=(804

) × (100 − 80

7 − 4)

(1007

)= 0.1126

Number of samples = (202

) × (806

) = 5.7095 × 1010

P(2 C in 8 S) =(202

) × (806

)

(1008

)= 0.3068



P(C > B) = P(8C ∩ 0B) + P(7C ∩ 1B) + P(6C ∩ 2B) + P(5C ∩ 3B)

=(208

) × (800

)

(1008

)+

(207

) × (801

)

(1008

)+

(206

) × (802

)

(1008

)

+(205

) × (803

)

(1008

)

= 0.0000 + 0.0000 + 0.0007 + 0.0068

= 0.0075

Problem (07): A box of 50 light bulbs contains 22 white light bulbs, 16 yellow light

bulbs and 12 red light bulbs. Suppose that 7 bulbs are randomly selected

for inspection.

(a) (2 points) What is the probability that five of the selected bulbs are

red.

(b) (2 points) What is the probability that four of the selected bulbs are

red, one is yellow and two are white.

(c) (1 points) Compute the probability that all selected bulbs have the

same colour.

(Question 3: (5 points) in Midterm Exam 2016)

Solution:

P(5R of 7) =(125

) × (22 + 16

2)

(22 + 16 + 12

7)

= 0.0056



P(4R ∩ 1Y ∩ 2W) =(124

) × (161

) × (222

)

(22 + 16 + 12

7)

= 0.0183

P(same colour) = P(7W) + P(7Y) + P(7R)

=(127

)

(22 + 16 + 12

7)

+(167

)

(22 + 16 + 12

7)

+(227

)

(22 + 16 + 12

7)

= 0.0017 + 0.0001 + 0.0000

= 0.0018



Section 3.4: The Poisson Distribution

Problem (01):

If X ∼ P(3.2), calculate:

(a) P(X = 1) (b) P(X ≤ 3)

(c) P(X ≥ 6) (d) P(X = 0 | X ≤ 3)

(e) E(X) (f) Var(X)


Solution:

P(X = 1) =e−𝜆 × 𝜆𝑥

𝑥!=

e−3.2 × (3.2)1

1!= 0.1304

≤

P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

= ∑e−𝜆 × 𝜆𝑥

𝑥!

=e−3.2 × (3.2)0

0!+

e−3.2 × (3.2)1

1!+

e−3.2 × (3.2)2

2!

+e−3.2 × (3.2)3

3!

= 0.6025

≥

P(X ≥ 6) = 1 − P(X ≤ 5)

= 1− [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

+ P(X = 4) + P(X = 5)]

= 1 − ∑e−𝜆 × 𝜆𝑥

𝑥!

= 0.1054



≤

P(X = 0 | X ≤ 3) =P((𝑋 = 0) ∩ (X ≤ 3))

P(𝑋 ≤ 3)

=P(X = 0)

P(X ≤ 3)

=(e−3.2 × (3.2)0

0! )

0.6025

= 0.0677

E(X) = 𝜆 = 3.2

Var(X) = 𝜆 = 3.2

Problem (02): The number of cracks in a ceramic tile has a Poisson distribution with a

mean of λ = 2.4.

(a) What is the probability that a tile has no cracks?

(b) What is the probability that a tile has four or more cracks?


Solution:

P(X = 0) =e−𝜆 × 𝜆𝑥

𝑥!=

e−2.4 × (2.4)0

0!= 0.0907

≥

P(X ≥ 4) = 1 − P(X ≤ 3)



= ∑e−𝜆 × 𝜆𝑥

𝑥!

= 1 −

[

e−2.4 × (2.4)0

0!+

e−2.4 × (2.4)1

1!

+e−2.4 × (2.4)2

2!+

e−2.4 × (2.4)3

3! ]

= 0.2213

Problem (03): On average there are about 25 imperfections in 100 meters of optical cable.

(a) Use the Poisson distribution to estimate the probability that there are

no imperfections in 1 meter of cable.

(b) What is the probability that there is no more than one imperfection

in 1 meter of cable?


Solution:

λ

P(X = 0) =e−𝜆 × 𝜆𝑥

𝑥!=

e−0.25 × (0.25)0

0!= 0.7788

≤

P(X ≤ 1) = P(X = 0) + P(X = 1)

= ∑e−𝜆 × 𝜆𝑥

𝑥!

=e−0.25 × (0.25)0

0!+

e−0.25 × (0.25)1

1!

= 0.9735



Problem (04): A box contains 500 electrical switches, each one of which has a probability

of 0.005 of being defective. Use the Poisson distribution to make an

approximate calculation of the probability that the box contains no more

than 3 defective switches.

Hint: Recall that the Poisson distribution with a parameter value of λ = np

can be used to approximate the B(n, p) distribution when n is very large

and the success probability p is very small.


Solution:

∼

λ ×

≤

P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

= ∑e−𝜆 × 𝜆𝑥

𝑥!

=e−2.5 × (2.5)0

0!+

e−2.5 × (2.5)1

1!+

e−2.5 × (2.5)2

2!

+e−2.5 × (2.5)3

3!

= 0.7576

Problem (05): In a scanning process, the number of misrecorded pieces of information

has a Poisson distribution with parameter λ = 9.2.

(a) What is the probability that there are between six and ten

misrecorded pieces of information?

(b) What is the probability that there are no more than four misrecorded

pieces of information?


Solution:



≤ ≤

P(6 ≤ X ≤ 10)

= P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9)

+ P(X = 10)

= ∑e−𝜆 × 𝜆𝑥

𝑥!

=e−9.2 × (9.2)6

6!+

e−9.2 × (9.2)7

7!+

e−9.2 × (9.2)8

8!

+e−9.2 × (9.2)9

9!+

e−9.2 × (9.2)10

10!

≤

P(X ≤ 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

+ P(X = 4)

= ∑e−𝜆 × 𝜆𝑥

𝑥!

=e−9.2 × (9.2)0

0!+

e−9.2 × (9.2)1

1!+

e−9.2 × (9.2)2

2!

+e−9.2 × (9.2)3

3!+

e−9.2 × (9.2)4

4!

Problem (06): Number of births in a hospital occurs randomly at an average rate of 1.8

births per hour. Let X be the number of births in a given hour and therefore

X ∼ P(1.8).

(a) What is the probability of observing 4 births in a given hour at the

hospital?



(b) What is the probability of observing more than or equal to 2 births

in a given hour at the hospital?

(From a previous exam)

Solution:

=

P(X = 4) =e−𝜆 × 𝜆𝑥

𝑥!=

e−1.8 × (1.8)4

4!= 0.0723

≥

P(X ≥ 2) = 1 − P(X ≤ 1) = 1 − [P(X = 0) + P(X = 1)]

= ∑e−𝜆 × 𝜆𝑥

𝑥!

=e−1.8 × (1.8)0

0!+

e−1.8 × (1.8)1

1!

= 0.4628

Problem (07): For the case of the thin copper wire, suppose that the number of flaws

follows a Poisson distribution with a mean of 2.3 flaws per millimetre.

(a) Determine the probability that there are exactly 2 flaws in 1

millimetre of wire.

(b) Determine the probability that there are 10 flaws in 5 millimetre of

wire.


Solution:

P(X = 2) =e−λ × λ𝑥

𝑥!=

e−2.3 × (2.3)2

2!= 0.2652



λ

λ

𝜆 =2.3 × 5.0

1.0= 11.50

P(X = 10) =e−λ × λ𝑥

𝑥!=

e−11.5 × (11.5)10

10!= 0.1129

Problem (08): A certain kind of sheet metal has on average, five defects per 10 square

feet. If a Poisson distribution has been assumed, what is the probability that

a 15 square feet sheet of the metal will have at least 3 defects?


Solution:

λ

λ

𝜆 =5.0 × 15.0

10.0= 7.50

P(X ≥ 3) = 1 − P(X ≤ 2)

= 1 − [𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2)]

= 1 − ∑e−λ × λ𝑥

𝑥!

= 1 − [e−7.5 × (7.5)0

0!+

e−7.5 × (7.5)1

1!+

e−7.5 × (7.5)2

2!]

= 0.9797

Problem (09): The historical records of rainstorm in a town indicate that, on the average,

there had been four rainstorms per year over the last 20 years. Assuming

that the occurrence of rainstorm follows a Poisson distribution.

(a) (5 points) What is the probability that there will be no rainstorms

next year?



(b) (5 points) What is the probability that there will be 2 or more

rainstorms in the next year?

(c) (5 points) What is the probability that there will be no rainstorm in

the next two years?

(Question 4: (15 points) in Final Exam 2010)

Solution:

P(X = 0) =e−λ × λ𝑥

𝑥!=

e−4 × (4)0

0!= 0.0183

≥

P(X ≥ 2) = 1 − P(X ≤ 1)

= 1 − [P(X = 0) + P(X = 1)]

= ∑e−λ × λ𝑥

𝑥!

= 1 − [e−4 × (4)0

0!+

e−4 × (4)1

1!]

= 0.9084

≥

λ

λ

𝜆 =4.0 × 2.0

1.0= 8.0

P(X = 0) =e−λ × λ𝑥

𝑥!=

e−8 × (8)0

0!= 0.0003

chapter 3: discrete probability distributions...

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