5 S1 Probability

Probability

• This Chapter is on Probability

• We will review methods used on GCSE Maths

• There will be new ideas introduced many of which will require diagrams to help solve

• Diagrams will include Probability Trees, Sample Spaces and Venn Diagrams

Probability

Probability Reminder• An experiment is a repeatable process that gives outcomes• An event is a collection of one (or more) outcomes• A sample space is the set of all possible outcomes in an

experiment

• P(event occurring) =

• Impossible event P = 0• Certain event P = 1

• Probabilities are only written as Fractions, Decimals and Percentages.

5A

No. outcomes in that eventTotal possible outcomes

Probability

Probability Reminder• Find the probability of a fair dice landing on a 5.

• P(event occurring) =

• P(5) =

5A

No. outcomes in that eventTotal possible outcomes

16

Probability

Probability Reminder• Two spinners are numbered 1-4. Both are spun and the sum

of the numbers calculated. Find P(5) and P(x > 5)

5A

5432

6543

7654

8765

1 2 3 4

1

2

3

4

Spinner 1

Spin

ner

2

Draw a sample space to show the outcomes.

P(5) =

P(x > 5) =

416

14

=

616

38

=

ProbabilityUsing Venn DiagramsVenn diagrams are a very useful way of representing Probabilities. They can also help you answer multi-part questions.

5B

A B

S

A rectangle labelled S represents the Sample Space

Circle B represents the Probability of

event B

Circle A represents the Probability of

event A

ProbabilityUsing Venn Diagrams

5B

A

A

A

B

B

B

The Area in the middle represents the Probability of A and B happening

together.P(A B)

The whole area represents the Probability of A or B happening (or them

together).P(A B)

‘n’ ‘and’

The area outside of A represents the Probability of A not happening.

P(A') = 1-P(A)

S

S

S

ProbabilityUsing Venn DiagramsA card is selected at random from a pack of 52 playing cards. Let A be the event that the card is an ace, and D be the event that the card is a diamond. Draw a Venn diagram to show this information.

5B

A D

S1) Always fill in the middle first. The

middle represents an ace and a diamond.

1 card

2) There are 4 aces in total, one of which has already been filled in

3 cards extra in ‘A’

3) There are 13 diamonds, one of which has been filled in

12 extra cards in ‘D’

4) 52 cards in total, subtract the 16 that have been used

36 cards left outside the circles

13 12

36

ProbabilityUsing Venn DiagramsA card is selected at random from a pack of 52 playing cards. Let A be the event that the card is an ace, and D be the event that the card is a diamond. Draw a Venn diagram to show this information.

5B

A D

S

13 12

36

P(A D) = 1 52

P(A D) = 16 52

4 13

P(A') = 48 52

12 13

P(A' D) = 12 52

3 13

‘Probability of an Ace and a Diamond’

‘Probability of an Ace or a Diamond’

‘Probability of it not being an Ace’

‘Probability of it not being an Ace, and being a Diamond’

ProbabilityUsing Venn DiagramsIn a class of 30 students, 7 are in the choir, 5 are in the school band and 2 are in both the choir and the band. Draw a Venn diagram to show this information.

5B

B C

S

1) Always fill in the middle first. The middle represents choir and band.

2 students

2) There are 5 students in the band, in total. 2 are already on the diagram.

3 students extra in ‘B’

3) There are 7 students in the choir, 2 of which are already on the diagram.

5 more students in ‘C’

4) 30 students in total, 10 already filled in.

20 students outside the circles

23 5

20

ProbabilityUsing Venn DiagramsIn a class of 30 students, 7 are in the choir, 5 are in the school band and 2 are in both the choir and the band. Draw a Venn diagram to show this information.

5B

B C

S

23 5

20

P(B') =

‘Probability of not being in the band’ 1 - P(B)

P(B') =5

1 - 30

P(B') =25

30

5

6

You could also have got 25/30 by counting the parts not in the ‘B’ circle.

ProbabilityUsing Venn DiagramsA vet surveys 100 clients. She finds out the following:25 have dogs 53 have cats 40 have fish15 have dogs and cats 10 have cats and fish 11 have dogs and fish7 have dogs, cats and fish

5B

D CS

7

F

8

34

356

26

1) Always fill in the middle first. The middle represents all 3 pets

2) Then fill in the parts where 2 circles overlap. Remember to take away the middle from each.

3) After this you can fill in the rest, based on what you have already completed

4) Remember to work out how many people have no pets (add up the numbers in the circle, and subtract from 100)

11

ProbabilityUsing Venn DiagramsA vet surveys 100 clients. She finds out the following:25 have dogs 53 have cats 40 have fish15 have dogs and cats 10 have cats and fish 11 have dogs and fish7 have dogs, cats and fish

5B

D CS

7

F

8

34

356

26

11

P(Dog only) = 6 100

P(Doesn't own Fish) = 60 100

3 5

P(None of these) = 11 100

3 50

Probability

Formulae from the Venn Diagram – The Addition Rule

5C

A B

S

i b - ia - i

1 – (a + b – i)

If P(A) = a and P(B) = b

And we let the intersection = iP(A B)

Then we can label a Venn diagram as follows:

The intersection i

The rest of A a – i

The rest of B b – i

The Area in the circles

(a – i) + i + (b – i)

a + b – i

So the remainder will be 1 – (a + b – i)

Probability

Formulae from the Venn Diagram – The Addition Rule

5C

A B

S

i b - ia - i

1 – (a + b – i)

If P(A) = a and P(B) = b

The Probability of A or B is the whole of the area inside the circles

P(A or B) = (a – i) + (b – i) + i

P(A or B) = a – i + b – i + i

P(A or B) = a + b - i

And we let the intersection = iP(A B)

P(A B) = P(A) + P(B) - P(A B)a b i-+P(A or B) =

P(A B) = P(A) + P(B) - P(A B) Rearranged you can also get this formula

Probability

Formulae from the Venn Diagram – The Addition RuleA and B are two events such that P(A) = 0.6, P(B) = 0.7 and P(A or B) = 0.9.Calculate: a)

b)c)d)

a)

5C

A B

SP(A B)P(A')

P(A' B)P(A' B)

P(A B) = P(A) + P(B) - P(A B)

P(A B) = 0.6 + 0.7 - 0.9

P(A B) = 0.4

0.4 0.30.2

0.1

Now you know the intersection, you can draw a Venn diagram!

Probability

Formulae from the Venn Diagram – The Addition RuleA and B are two events such that P(A) = 0.6, P(B) = 0.7 and P(A or B) = 0.9.Calculate: b)

c)d)

b)

c)

d)

5C

A B

SP(A')

P(A' B)P(A' B)

0.4 0.30.2

0.1

P(A') = 0.4

P(A' B) = 0.8

P(A' B) = 0.3

‘Probability of not A’

‘Probability of not A, or B’

‘Probability of not A, and B’

Probability

Formulae from the Venn Diagram – The Multiplication RuleConditional Probability is where the probability of an event is affected by whether another event has already occurred or not.

For example, the probability of choosing an ace from a pack of cards will be affected if a random card has been removed

We have learnt the ‘addition’ rule, and now we will learn the ‘multiplication rule’

The multiplication rule will allow us to solve problems involving conditional probability

5D

Probability

Formulae from the Venn Diagram – The Multiplication RuleTwo fair spinners are both numbered 1-4. They are thrown together and the sum recorded. Given that at least one spinner lands on a 3, find the probability of the spinners indicating a sum of 5.

5D

5432

6543

7654

8765

1 2 3 4

1

2

3

4

Spinner 1

Spin

ner

2

Using the sample space:

We only consider the results that have a 3 as one of the numbers (there are 7 possibilities here)

Out of these, 2 have a sum of 5

So P(sum of 5 | at least one 3)

2/7

Probability

Formulae from the Venn Diagram – The Multiplication RuleWe are going to work out the formula for the probability of event B, given that event A has happened.

5D

A B

i b - ia - i

We are saying that A has happened, and so are only including the ‘A’

area

‘i’ is the part of the circle where B has

happened

As A has happened, the ‘A’ area represents all the

possibilities

So the probability B happens, given A has:

i

a i i i

a

( )( | )

( )

P B AP B A

P A

The vertical line means ‘given’

Probability

Formulae from the Venn Diagram – The Multiplication RuleWe are going to work out the formula for the probability of event B, given that event A has happened.

5D

A B

i b - ia - i

( )( | )

( )

P B AP B A

P A

( ) ( | ) ( )P B A P B A P A

Re-arranging, we get…

( ) ( | ) ( )P A B P A B P B

OR

ProbabilityFormulae from the Venn Diagram – The Multiplication RuleC and D are two events such that P(C) = 0.2, P(D) = 0.6 and P(C|D) = 0.3.

Find: a) b) c)P(D | C) P(C' D') P(C' D)

We will construct a Venn Diagram to help…

We need the probability of the intersection:

( ) ( | ) ( )P C D P C D P D

( ) 0.3 0.6P C D

( ) 0.18P C D

C D

0.18

0.02

0.42

P(C) = 0.2, so subtract

0.18

P(D) = 0.6, so subtract

0.18

0.38

The probabilities must add up to 1

( )( | )

( )

P B AP B A

P A

( ) ( | ) ( )P B A P B A P A

5D



C D

0.18

0.02

0.42

0.38

P(D C)P(D | C) =

( )P C

0.18P(D | C) =

0.2

P(D | C) = 0.9

a)

( )( | )

( )

P B AP B A

P A

( ) ( | ) ( )P B A P B A P A

5D



C D

0.18

0.02

0.42

0.38

b)

( )( | )

( )

P B AP B A

P A

( ) ( | ) ( )P B A P B A P A

0.9

P(C' D') =

‘Probability of not C and not D’

c)

‘Probability of not C and D’

P(C' D) =

0.38

0.42

0.38

0.42

5D

ProbabilityFormulae from the Venn Diagram – The Multiplication RuleLet A and B be events such that P(A) = 3/10, P(B) = 2/5 and = 1/2.

Find: a) b)P(B | A)We will construct a Venn Diagram to help…

We need the probability of the intersection:

( ) ( ) ( ) ( )P B A P B P A P B A 2 3 1

( )5 10 2

P B A

A B

1 5

P(B A)

4 3 5( )

10 10 10P B A

1( )

5P B A

1 10

1 5

1 2

3/10 – 1/5

3/10 – 2/10

2/5 – 1/5

1 - (1/10 + 1/5 + 1/5)

1 - (1/10 + 2/10 + 2/10)

P(A) = 3/10 in total

P(B) = 2/5 in total

1 - (5/10)

Denominators must be the

same

5D

P(B | A')

( )( | )

( )

P B AP B A

P A

( ) ( | ) ( )P B A P B A P A

P(A B) = P(A) + P(B) - P(A B)


Find: a) b)P(B | A)

A B

1 5

P(B A)

1 10

1 5

1 2

15P(B | A) =

310

a)P(B A)

P(B | A) = P(A)

2P(B | A) =

3

2 3

5D

P(B | A')

( )( | )

( )

P B AP B A

P A

( ) ( | ) ( )P B A P B A P A

P(A B) = P(A) + P(B) - P(A B)


Find: a) b)P(B | A) P(B | A')

A B

1 5

P(B A)

1 10

1 5

1 2

15P(B | A') =

710

b)P(B A')

P(B | A') = P(A')

2P(B | A') =

7

2 3

P(A’) = 1 – P(A)

= 1 – 3/10

P(B A')

5D

( )( | )

( )

P B AP B A

P A

( ) ( | ) ( )P B A P B A P A

P(A B) = P(A) + P(B) - P(A B)

ProbabilityTree Diagrams

You will have seen Tree Diagrams at GCSE level, and they can also be used to represent conditional probabilities.

The number of spectators at an event is dependent on the weather. On a rainy day, the probability of a big turnout is 0.4. However, if it does not rain, there is a probability of 0.9 that there will be a big turnout. The weather forecast gives a 0.75 probability of rain. Show this on a tree diagram.

5E

R

R’

B

B

B’

B’

0.75

0.25

0.1

0.9

0.4

0.6

The rain is the independent event (ie – it affects the probability of the other), so it comes first.

P(R) = 0.75

P(R’) = 0.25The second set of possibilities are a high turnout (B), or not (B’)

The probabilities are different depending on whether it rained or not…


5E

R

R’

B

B

B’

B’

0.75

0.25

0.1

0.9

0.4

0.6

You can use the Multiplication rule to work out probabilities.

( )P B R ‘Probability of a big turnout and rain’

( | ) ( )P B R P R( )P B R 0.4 0.75( )P B R 0.3

( )P B R 0.3


5E

R

R’

B

B

B’

B’

0.75

0.25

0.1

0.9

0.4

0.6


( ' )P B R ‘Probability of not a big turnout and rain’ ( ' | ) ( )P B R P R

( ' )P B R 0.6 0.75( ' )P B R 0.3

( )P B R 0.3

( ' )P B R 0.45


5E

R

R’

B

B

B’

B’

0.75

0.25

0.1

0.9

0.4

0.6


( ')P B R ‘Probability of a big turnout and not rain’ ( | ') ( ')P B R P R

( ')P B R 0.9 0.25( ')P B R 0.225

( )P B R 0.3

( ' )P B R 0.45

( ')P B R 0.225


5E

R

R’

B

B

B’

B’

0.75

0.25

0.1

0.9

0.4

0.6


( ' ')P B R ‘Probability of not a big turnout and not rain’ ( ' | ') ( ')P B R P R

( ' ')P B R 0.1 0.25( ' ')P B R 0.225

( )P B R 0.3

( ' )P B R 0.45

( ')P B R 0.225

( ')P B R 0.025

Basically, remember to multiply along

each ‘path’

Checking…

0.3 + 0.45 + 0.225 + 0.025

1


5E

R

R’

B

B

B’

B’

0.75

0.25

0.1

0.9

0.4

0.6

Calculate the probability of a big turnout.

( )P B R 0.3

( ' )P B R 0.45

( ')P B R 0.225

( ' ')P B R 0.025

( ) ( ')P B R P B R 0.3 0.225

( )P B ( )P B

0.525( )P B

There are 2 ways of having a big turnout


A bag contains 7 green beads and 5 blue beads. A bead is taken at random, the colour recorded and the bead is not replaced. A second is then taken and the colour recorded. Find P(1 Green and 1 Blue).

5E

G1

B1

G2

G2

B2

B2

7 12

There is the possibility of Green or Blue both times.

P(G1) = 7/12

P(B1) = 5/12

The second set of possibilities depend on what colour was taken the first time. There will be 11 left, and one less of either Green or Blue.

5 12

4 11

7 11

5 11

6 11

One less Green

Blue the same as to begin with

One less Blue

Green the same as to begin with


A bag contains 7 green beads and 5 blue beads. A bead is taken at random, the colour recorded and the bead is not replaced. A second is then taken and the colour recorded. Find P(1 Green and 1 Blue).

5E

G1

B1

G2

G2

B2

B2

7 12

5 12

4 11

7 11

5 11

6 11

As we want one of each, there are 2 possible routes:

( )P G B 35 35

( )132 132

P G B

7 12

5 11

35 132

x =

5 12

7 11

35 132

x =70

( )132

P G B

35( )

66P G B

ProbabilityMutually Exclusive and Independent EventsWhen 2 events cannot happen at the same time, they are Mutually Exclusive.

If we apply this to the Addition Rule:

You can also work backwards. If the above is true then the events are Mutually Exclusive.

5F

A B

S

( ) 0P A B

( )P A B ( )P A ( )P B ( )P A B

( )P A B ( )P A ( )P BThe Addition Rule for

Mutually exclusive Events

ProbabilityMutually Exclusive and Independent EventsWhen one event has no effect on another, they are said to be independent.

So the Probability of A, given B, is just the same as the probability of A on its own.

Applying this to the Multiplication Rule.

Again, you can work backwards. If you put the numbers you are given into the above formula, and it works, then the events are independent.

5F

( | )P A B ( )P A

( )P A B ( | )P A B ( )P B

( )P A B ( )P A ( )P BThe Multiplication Rule for

Independent Events

ProbabilityMutually Exclusive and Independent EventsEvents A and B are Mutually Exclusive and P(A) = 0.2 and P(B) = 0.4

Calculate:a)

b)

c)

5F

( )P A B

( ')P A B

( ' ')P A B

A B

S

0.2

0.4

0.4

( ) ( )P A P B 0.2 0.4 0.6

( )P A

1 ( )P A B

0.2

1 0.6 0.4

Construct a Venn Diagram

Mutually Exclusive, so the circles are separate

ProbabilityMutually Exclusive and Independent EventsEvents C and D are Independent and P(C) = 1/3 and P(D) = 1/5

Calculate:a)

b)

c)

5F

( )P C D

( ')P C D

( ' ')P C D

C D

S

4/152/15

8/15

( ) ( )P C P D 1 1

3 5

Construct a Venn Diagram

1

15

1/15

1 1 -

3 15

5 1 -

15 15

1 1 -

5 15

3 1 -

15 15

P(C) = 1/3 in total so:

P(D) = 1/3 in total so:

4

15

8

15

Summary• We have now finished all the topics for Probability

• You must remember both the Addition Rule and the Multiplication Rule. If you aren’t sure, think ‘what am I trying to find out?’

• The best way to solve a problem is to draw a diagram to help, when you have enough information to do so

• You also need to remember how the rules vary for Mutually Exclusive and Independent Events

5 S1 Probability

Documents