joan ridgway. all probabilities lie somewhere on a scale between “ impossible ” and “ certain...

Joan Ridgway

All probabilities lie somewhere on a

scale between “Impossible” and

“Certain”

Impo

ssib

le C

erta

in

Win

ning

the

lott

ery

Get

ting

a C

for

GC

SE M

aths

It w

ill r

ain

tom

orro

w

Thro

win

g a

6 on

a d

ice

Toss

ing

a co

in a

nd

…ge

ttin

g “h

eads

”

All probabilities lie somewhere on a

scale between “Impossible” and

“Certain”

Impo

ssib

le C

erta

in

Win

ning

the

lott

ery

Get

ting

a C

for

GC

SE M

aths

Toss

ing

a co

in a

nd

…ge

ttin

g “h

eads

”

Thro

win

g a

6 on

a d

ice

The probability scale goes from 0 to

1

0 1

It w

ill r

ain

tom

orro

w

Impo

ssib

le C

erta

in

Win

ning

the

lott

ery

Get

ting

a C

for

GC

SE

Mat

hs

Thro

win

g a

6 on

a d

ice

Probabilities can be expressed either as fractions or as decimals (and sometimes as

percentages)

0 10.5 95% 1 14,000,000

16

It w

ill r

ain

tom

orro

w

Toss

ing

a co

in a

nd

…ge

ttin

g “h

eads

”

0.7

The probability of throwing a

6 with a fair dice is 16

P(6) = 16

So the probability of not

throwing a 6 is 56

P(not 6) = 1- 16 = 5

6

If the probability that it will rain tomorrow is 0.7 …..

P(rain) = 0.7

Then the probability that it

will not rain tomorrow is 0.3

P(not rain) = 1 – 0.7 = 0.3

Suppose I toss a coin:

What is the probability of getting a head?

0.5, ½ or 50%

What is the probability of getting two heads?

If I toss the coin twice, I would get one of these combinations:

Heads, Heads Heads, Tails Tails, Heads Tails, Tails

or

H, H H, T T, H T, T

Only one of these four combinations is two heads

Suppose I toss two coins:

What is the probability of getting two heads?

If I toss the coin twice, I would get one of these combinations:

Heads, Heads Heads, Tails Tails, Heads Tails,Tails

or

H, H H, T T, H T, T

Only one of these four combinations is two heads

H H

So the probability of getting a two heads in a row is

¼

Suppose I toss two coins:

A Sample Space is a list of all the possible outcomes, e.g. HH, HT, TH, TT We can show this in a Sample Space Diagram:

There are 4 possible outcomes if you toss a coin twiceSo the probability of two heads is ¼

T, TT, HT

H, TH, HH

TH

Second CoinF

irst

Coi

n

Suppose I throw a die.

There are 6 equally likely outcomes.

Suppose I throw two dice.

Suppose I throw two dice.

We can show the possible outcomes in a Sample Space Diagram:

There are 36 (6 x 6) possible outcomes if you throw two dice.

6, 66, 56, 46, 36, 26, 16

5, 65, 55, 45, 35, 25, 15

4, 64, 54, 44, 34, 24, 14

3, 63, 53, 43, 33, 23, 13

2, 62, 52, 42, 32, 22, 12

1, 61, 51, 41, 31, 21, 11

654321

Second DiceF

irst

Dic

e

If you throw two dice, what is the probability of getting a “double”?

6, 66, 56, 46, 36, 26, 16

5, 65, 55, 45, 35, 25, 15

4, 64, 54, 44, 34, 24, 14

3, 63, 53, 43, 33, 23, 13

2, 62, 52, 42, 32, 22, 12

1, 61, 51, 41, 31, 21, 11

654321

Second DiceF

irst

Dic

e

6 out of the 36 possible outcomes are “doubles”, so the probability is 6

36

6, 66, 56, 46, 36, 26, 16

5, 65, 55, 45, 35, 25, 15

4, 64, 54, 44, 34, 24, 14

3, 63, 53, 43, 33, 23, 13

2, 62, 52, 42, 32, 22, 12

1, 61, 51, 41, 31, 21, 11

654321

Second DiceF

irst

Dic

e


6 out of the 36 possible outcomes are “doubles”, so the probability is1

6

6, 66, 56, 46, 36, 26, 16

5, 65, 55, 45, 35, 25, 15

4, 64, 54, 44, 34, 24, 14

3, 63, 53, 43, 33, 23, 13

2, 62, 52, 42, 32, 22, 12

1, 61, 51, 41, 31, 21, 11

654321

Second DiceF

irst

Dic

e


What is the probability of scoring 9 or more?

6, 66, 56, 46, 36, 26, 16

5, 65, 55, 45, 35, 25, 15

4, 64, 54, 44, 34, 24, 14

3, 63, 53, 43, 33, 23, 13

2, 62, 52, 42, 32, 22, 12

1, 61, 51, 41, 31, 21, 11

654321

Second DiceF

irst

Dic

e

10 out of the 36 possible outcomes add up to 9 or more, so the probability is10

36

6, 66, 56, 46, 36, 26, 16

5, 65, 55, 45, 35, 25, 15

4, 64, 54, 44, 34, 24, 14

3, 63, 53, 43, 33, 23, 13

2, 62, 52, 42, 32, 22, 12

1, 61, 51, 41, 31, 21, 11

654321

Second DiceF

irst

Dic

e


10 out of the 36 possible outcomes add up to 9 or more, so the probability is

6, 66, 56, 46, 36, 26, 16

5, 65, 55, 45, 35, 25, 15

4, 64, 54, 44, 34, 24, 14

3, 63, 53, 43, 33, 23, 13

2, 62, 52, 42, 32, 22, 12

1, 61, 51, 41, 31, 21, 11

654321

Second DiceF

irst

Dic

e


5 18

We cannot always calculate the probability of an event; sometimes we have to estimate it.

Suppose we did not know whether a dice was fair or weighted. We could throw it 100 times to find how often we threw a 6.

We would expect to get a 6 about once in every 6 throws, as the probability should be 1 in 6.

We cannot always calculate the probability of an event; sometimes we have to estimate it.

Suppose we did not know whether a dice was fair or weighted. We could throw it 100 times to find how often we threw a 6.

We threw 35 sixes out of a total of 100 throws

6

5

4

3

2

1

35 llll llll llll llll llll llll llll12 llll llll ll13 llll llll lll13 llll llll lll 12 llll llll ll15 llll llll llll

FrequencyTally

100


The relative frequency of throwing a 6 with this dodgy dice is:

35 100


The relative frequency of throwing a 6 with this dodgy dice is:

The probability of throwing a 6 with a normal dice is:

or 0.35

1 6 or 0.16

.

We threw a 6 more than twice as often as would be expected.The dice is probably weighted!!!

35 100

Go back to the coins.

Remember there were 4 possible outcomes if I toss 2 coins

Heads, Heads Heads, Tails Tails, Heads Tails,Tails

or

There are 4 possible outcomes because 2 x 2 = 4, just as for two dice there are 36 possible outcomes because 6 x 6 = 36

H H H T T H T T

If I toss three coins, what are the possible combinations?

T T T

H H H

H H T H T H T H H

H T T T H T T T H

– 3 Heads

2 Heads (1 Tail)

1 Head (2 Tails)

– 0 Heads (3 Tails)There are 8 possible outcomes because

2 x 2 x 2 = 8


The probability of 3 Heads is 1 8


The probability of 1 Head is 3 8


T T T

H H H

H H T H T H T H H

H T T T H T T T H

– 3 Heads

2 Heads (1 Tail)

1 Head (2 Tails)

– 0 Heads (3 Tails)


T T T

H H H H H T H T H T H H H T T T H T T T H

– 3 Heads

2 Heads (1 Tail)

1 Head (2 Tails)

– 0 Heads (3 Tails)



The probability of 1 Head is 3 8

The probability of 0 Heads is 1 8Simulation Hyperlink

http://www.mathsonline.co.uk/nonmembers/resource/prob/coins.html

First coin Second coin

Heads

Tails

1 .2

1 .2

Heads

Heads

Tails

Tails

1 .2

1 .2

1 .2

1 .2

P(HH) =

P(HT) =

P(TH) =

P(TT) =

4

1

2

1

2

1

4

1

2

1

2

1

4

1

2

1

2

1

4

1

2

1

2

1

Check that the probabilities add up to 1. 14

1

4

1

4

1

4

1

Sometimes it is helpful to draw tree diagrams.

If I toss a coin twice, what is the probability of getting at least one head?

The Watsons regard one boy and one girl as the ideal family. What is the chance of getting one boy and one girl in their planned family of two?

G

B

G

G

B

B

First Child Second Child Combined

0.5

0.5

0.5

0.5

0.5

0.5 P(G,G) = 0.5 x 0.5 = 0.25

P(B,B) = 0.5 x 0.5 = 0.25

P(B,G) = 0.5 x 0.5 = 0.25

P(G,B) = 0.5 x 0.5 = 0.25

The probability of getting one of each is:

P(G,B) + P(B,G) = 0.25 + 0.25 = 0.5

Total = 1.00


Joan travels to work on her bicycle. She has to go through two sets of traffic lights on her way. At the first set of lights, the probability that they will be green is 7/10. At the second set the probability that they will be green is 3/5

First set Second set

Green

Not Green

7 10

3 10

Green

Green

Not Green

Not Green

3 .5

3 .5

2 .5

2 .5

P(both green) =

P(green, not green) =

P(not green, green) =

P(both not green) =

50

21

5

3

10

7

50

14

5

2

10

7

50

9

5

3

10

3

50

6

5

2

10

3

Check that the probabilities add up to 1. 150

50

50

6

50

9

50

14

50

21



1) What is that probability that both sets will be green?

2) What is the probability that she will have to stop once and only once at a set of lights?

You should simplify the probabilities where you can.

25

7

25

3





Green

Not Green

Green

Green

Not Green

Not Green


7 10

3 .5

3 .5

2 .5

3 10

2 .5

P(both green) =



P(both not green) =

50

21

5

3

10

7

50

14

5

2

10

7

50

9

5

3

10

3

50

6

5

2

10

3

1) P(both green) =50

21





Green

Not Green

Green

Green

Not Green

Not Green


7 10

3 .5

3 .5

2 .5

3 10

2 .5

P(both green) =



P(both not green) =

25

7

25

3

50

21

5

3

10

7

50

14

5

2

10

7

50

9

5

3

10

3

50

6

5

2

10

3

2) P(only stopping once) =50

14





Green

Not Green

Green

Green

Not Green

Not Green


7 10

3 .5

3 .5

2 .5

3 10

2 .5

50

9

50

23

P(both green) =



P(both not green) =

25

7

25

3

50

21

5

3

10

7

50

14

5

2

10

7

50

9

5

3

10

3

50

6

5

2

10

3

Two events are independent if they have no effect on each other.

If you choose two items with replacement the events will be independent because the second choice will not be affected by the first choice.

Independent and Dependent Events

There are 10 crayons in a bag; 5 blue ones, 3 red ones and 2 green ones. I pick one at random, then put it back and choose another.

Draw a tree diagram to show the probabilities.

First Crayon Second Crayon

Blue

Red

Green

Blue

Red

Green

Blue

Red

Green

Blue

Red

Green

3 10

5 10

2 10

5 10 3

10

2 10

5 10 3

10

2 10

5 10 3

10

2 10


Blue

Red

Green

Blue

Red

Green

Blue

Red

Green

Blue

Red

Green

3 10

5 10

2 10

5 10 3

10

2 10

5 10 3

10

2 10

5 10 3

10

2 10

What is the probability that both crayons will be green?

100

4

10

2

10

2

25

1


Blue

Red

Green

Blue

Red

Green

Blue

Red

Green

Blue

Red

Green

3 10

5 10

2 10

5 10 3

10

2 10

5 10 3

10

2 10

5 10 3

10

2 10

What is the probability that I will choose one blue crayon and one red crayon? (in either order)

100

15

10

5

10

3

100

15

10

3

10

5

or


Blue

Red

Green

Blue

Red

Green

Blue

Red

Green

Blue

Red

Green

3 10

5 10

2 10

5 10 3

10

2 10

5 10 3

10

2 10

5 10 3

10

2 10

What is the probability that I will choose one blue crayon and one red crayon? (in either order)

100

15

10

5

10

3

100

15

10

3

10

5

or

100

30

100

15

100

15

10

3

Two events are dependent if one event will affect the other. The probability is said to be conditional.

If you choose two items without replacement the events will be dependent because the first item you choose cannot be chosen again.

Independent and Dependent Events

There are 10 jellybabies in a bag; 5 red ones, 3 green ones and 2 yellow ones. I pick one at random and eat it, and then I choose another.

Draw a tree diagram to show the probabilities.

First Jellybaby Second Jellybaby

Red

Green

Yellow

Red

Green

Yellow

Red

Green

Yellow

Red

Green

Yellow

3 10

5 10

2 10


Red

Green

Yellow

Red

Green

Yellow

Red

Green

Yellow

Red

Green

Yellow

3 10

5 10

2 10

4 9

If I choose a red one first and eat it, there will be only 9 jellybabies left, and only 4 will be red ones!!

BE CAREFUL!!

3 9

2 9


Red

Green

Yellow

Red

Green

Yellow

Red

Green

Yellow

Red

Green

Yellow

3 10

5 10

2 10

4 9

3 9

2 9


Red

Green

Yellow

Red

Green

Yellow

Red

Green

Yellow

Red

Green

Yellow

3 10

5 10

2 10

4 9

If I choose a green one first and eat it, there will be only 9 jellybabies left, and only 2 will be green ones!!

BE CAREFUL!!

3 9

2 95 9

2 9

2 9


Red

Green

Yellow

Red

Green

Yellow

Red

Green

Yellow

Red

Green

Yellow

3 10

5 10

2 10

4 9

3 9

2 95 9

2 9

2 9


Red

Green

Yellow

Red

Green

Yellow

Red

Green

Yellow

Red

Green

Yellow

3 10

5 10

2 10

4 9

If I choose a yellow one first and eat it, there will be only 9 jellybabies left, and only 1 will be a yellow one!!

BE CAREFUL!!

3 9

2 95 9

2 9

2 9 5

9

3 9

1 9

What is the probability that both jellybabies will be yellow?

90

2

9

1

10

2

45

1


Red

Green

Yellow

Red

Green

Yellow

Red

Green

Yellow

Red

Green

Yellow

3 10

5 10

2 10

4 9

3 9

2 95 9

2 9

2 9 5

9

3 9

1 9

What is the probability that I will choose (and eat) one red jelly baby and one green jelly baby? (in either order)

90

15

9

5

10

3


Red

Green

Yellow

Red

Green

Yellow

Red

Green

Yellow

Red

Green

Yellow

3 10

5 10

2 10

4 9

3 9

2 95 9

2 9

2 9 5

9

3 9

1 9

90

15

9

3

10

5

or

What is the probability that I will choose (and eat) one red jelly baby and one green jelly baby? (in either order)

90

15

9

5

10

3


Red

Green

Yellow

Red

Green

Yellow

Red

Green

Yellow

Red

Green

Yellow

3 10

5 10

2 10

4 9

3 9

2 95 9

2 9

2 9 5

9

3 9

1 9

90

15

9

3

10

5

or

90

30

90

15

90

15

3

1

joan ridgway. all probabilities lie somewhere on a scale between “ impossible ” and “ certain...

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