1. Probability

2. 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

3. Teachings for Exercise 5A

4. 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.
No. outcomes in that event
Total possible outcomes 5A

5. Probability Probability Reminder
Find the probability of a fair dice landing on a 5.
P(event occurring) =
P(5) =
No. outcomes in that event
Total possible outcomes 1 6 5A

6. 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) 5 6 7 8 4
Draw a sample space to show the outcomes.
P(5) =
P(x > 5) = 4 5 6 7 3
Spinner 2 4 1 = 3 4 5 6 2 16 4 2 3 4 5 1 6 3 1 2 3 4 = 16 8
Spinner 1 5A

7. Teachings for Exercise 5B

8. Probability Using Venn Diagrams
Venn diagrams are a very useful way of representing Probabilities. They can also help you answer multi-part questions. S A B
A rectangle labelled S represents the Sample Space
Circle A represents the Probability of event A
Circle B represents the Probability of event B 5B

9. The area outside of A represents the Probability of A not happening.
Using Venn Diagrams A B
The Area in the middle represents the Probability of A and B happening together.
‘n’  ‘and’ S
The whole area represents the Probability of A or B happening (or them together). A B S
The area outside of A represents the Probability of A not happening. A B 5B

10. Probability Using Venn Diagrams
A 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.
Always fill in the middle first. The middle represents an ace and a diamond.
 1 card
There are 4 aces in total, one of which has already been filled in
 3 cards extra in ‘A’
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 S A D 3 12 1 36 5B

11. Probability Using Venn Diagrams
A 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.
1 52 S
‘Probability of an Ace and a Diamond’
16 52
4 13 A D
‘Probability of an Ace or a Diamond’ 3 12 1
48 52
12 13
‘Probability of it not being an Ace’ 36
12 52
3 13
‘Probability of it not being an Ace, and being a Diamond’ 5B

12. Probability Using Venn Diagrams
In 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.
Always fill in the middle first. The middle represents choir and band.
 2 students
There are 5 students in the band, in total. 2 are already on the diagram.
 3 students extra in ‘B’
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 S B C 3 5 2 20 5B

13. Probability Using Venn Diagrams
In 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.
‘Probability of not being in the band’ S B C 3 5 2 20
You could also have got 25/30 by counting the parts not in the ‘B’ circle. 5B

14. Probability Using Venn Diagrams
A vet surveys 100 clients. She finds out the following:
25 have dogs 53 have cats 40 have fish
15 have dogs and cats 10 have cats and fish 11 have dogs and fish
7 have dogs, cats and fish S
Always fill in the middle first. The middle represents all 3 pets D C 6 8 35
2) Then fill in the parts where 2 circles overlap. Remember to take away the middle from each. 7 4 3
3) After this you can fill in the rest, based on what you have already completed 26
4) Remember to work out how many people have no pets (add up the numbers in the circle, and subtract from 100) F 11 5B

15. Probability Using Venn Diagrams
A vet surveys 100 clients. She finds out the following:
25 have dogs 53 have cats 40 have fish
15 have dogs and cats 10 have cats and fish 11 have dogs and fish
7 have dogs, cats and fish S
6 100
3 50 D C 6 8 35
60 100
3 5 7 4 3
11 100 26 F 11 5B

16. Teachings for Exercise 5C

17. Probability Formulae from the Venn Diagram – The Addition Rule S A B
If P(A) = a and P(B) = b S
And we let the intersection = i
Then we can label a Venn diagram as follows: A B
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)
a - i i
b - i
1 – (a + b – i) 5C

18. Rearranged you can also get this formula
Probability
Formulae from the Venn Diagram – The Addition Rule
If P(A) = a and P(B) = b S
And we let the intersection = i A 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
a - i i
b - i
1 – (a + b – i)
P(A or B) = a + b - i
Rearranged you can also get this formula 5C

19. Probability Formulae from the Venn Diagram – The Addition Rule
A 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) S A B
0.2
0.4
0.3
0.1
Now you know the intersection, you can draw a Venn diagram! 5C

20. Probability Formulae from the Venn Diagram – The Addition Rule
A 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) S A B
0.2
0.4
0.3
‘Probability of not A’
0.1
‘Probability of not A, or B’
‘Probability of not A, and B’ 5C

21. Teachings for Exercise 5D

22. Probability Formulae from the Venn Diagram – The Multiplication Rule
Conditional 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

23. Probability Formulae from the Venn Diagram – The Multiplication Rule 4
Two 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.
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 5 6 7 8 4 4 5 6 7 3
Spinner 2 3 4 5 6 2 2 3 4 5 1 1 2 3 4
Spinner 1 5D

24. Probability Formulae from the Venn Diagram – The Multiplication Rule A
We are going to work out the formula for the probability of event B, given that event A has happened.
We are saying that A has happened, and so are only including the ‘A’ area A B
a - i i
b - i
As A has happened, the ‘A’ area represents all the possibilities
‘i’ is the part of the circle where B has happened
So the probability B happens, given A has:
The vertical line means ‘given’ 5D

25. Probability Formulae from the Venn Diagram – The Multiplication Rule A
We are going to work out the formula for the probability of event B, given that event A has happened. A B
a - i i
b - i
Re-arranging, we get… OR 5D

26. The probabilities must add up to 1
Probability
Formulae from the Venn Diagram – The Multiplication Rule
C 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) = 0.6, so subtract 0.18
We will construct a Venn Diagram to help…
We need the probability of the intersection: C D
0.02
0.18
0.42
0.38
P(C) = 0.2, so subtract 0.18
The probabilities must add up to 1 5D

27. Probability
Formulae from the Venn Diagram – The Multiplication Rule
C 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) a) C D
0.02
0.18
0.42
0.38 5D

28. Probability
Formulae from the Venn Diagram – The Multiplication Rule
C 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)
0.9
0.38
0.42 b) C D
‘Probability of not C and not D’
0.02
0.18
0.42 c)
‘Probability of not C and D’
0.38 5D

29. Denominators must be the same
Probability
Formulae from the Venn Diagram – The Multiplication Rule
Let A and B be events such that P(A) = 3/10, P(B) = 2/5 and = 1/2.
Find: a) b)
P(B) = 2/5 in total
We will construct a Venn Diagram to help…
We need the probability of the intersection:
2/5 – 1/5 A B
1 10
1 5
1 5
Denominators must be the same
1 2
P(A) = 3/10 in total
1 - (1/10 + 1/5 + 1/5)
1 - (1/10 + 2/10 + 2/10)
3/10 – 1/5
1 - (5/10)
3/10 – 2/10 5D

30. Probability
Formulae from the Venn Diagram – The Multiplication Rule
Let A and B be events such that P(A) = 3/10, P(B) = 2/5 and = 1/2.
Find: a) b)
2 3 a) A B
1 10
1 5
1 5
1 2 5D

31. Probability
Formulae from the Venn Diagram – The Multiplication Rule
Let A and B be events such that P(A) = 3/10, P(B) = 2/5 and = 1/2.
Find: a) b)
2 3 b) A B
1 10
1 5
1 5
1 2
P(A’) = 1 – P(A)
= 1 – 3/10 5D

32. Teachings for Exercise 5E

33. Probability
Tree 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.
The rain is the independent event (ie – it affects the probability of the other), so it comes first.
P(R) = 0.75
P(R’) = 0.25 B
0.4 R
0.75 B’
0.6 B
0.9
The second set of possibilities are a high turnout (B), or not (B’)
 The probabilities are different depending on whether it rained or not…
0.25 R’
0.1 B’ 5E

34. Probability B R B’ B R’ B’ 5E Tree Diagrams 0.4 0.75 0.6 0.9 0.25 0.1
You can use the Multiplication rule to work out probabilities.
‘Probability of a big turnout and rain’ 5E

35. Probability B R B’ B R’ B’ 5E Tree Diagrams 0.4 0.75 0.6 0.9 0.25 0.1
You can use the Multiplication rule to work out probabilities.
‘Probability of not a big turnout and rain’ 5E

36. Probability B R B’ B R’ B’ 5E Tree Diagrams 0.4 0.75 0.6 0.9 0.25 0.1
You can use the Multiplication rule to work out probabilities.
‘Probability of a big turnout and not rain’ 5E

37. Basically, remember to multiply along each ‘path’
Probability
Tree Diagrams B
0.4
Checking… 1 R
0.75 B’
0.6 B
0.9
0.25 R’
0.1 B’
You can use the Multiplication rule to work out probabilities.
‘Probability of not a big turnout and not rain’
Basically, remember to multiply along each ‘path’ 5E

38. There are 2 ways of having a big turnout
Probability
Tree Diagrams B
0.4 R
0.75 B’
0.6 B
0.9
0.25 R’
0.1 B’
Calculate the probability of a big turnout.
There are 2 ways of having a big turnout 5E

39. Probability
Tree Diagrams
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).
One less Green
6 11
There is the possibility of Green or Blue both times.
P(G1) = 7/12
P(B1) = 5/12
Blue the same as to begin with G2 G1
7 12
5 11 B2
7 11 G2
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. B1
4 11 B2
Green the same as to begin with
One less Blue 5E

40. Probability
Tree Diagrams
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).
6 11
As we want one of each, there are 2 possible routes: G2 G1
7 12
7 12
5 11
35 132
5 11 B2 x =
7 11
5 12
7 11
35 132 G2 x =
5 12 B1
4 11 B2 5E

41. Teachings for Exercise 5F

42. The Addition Rule for Mutually exclusive Events
Probability S
Mutually Exclusive and Independent Events
When 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. A B
The Addition Rule for Mutually exclusive Events 5F

43. The Multiplication Rule for Independent Events
Probability
Mutually Exclusive and Independent Events
When 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.
The Multiplication Rule for Independent Events 5F

44. Probability
Mutually Exclusive and Independent Events
Events A and B are Mutually Exclusive and P(A) = 0.2 and P(B) = 0.4
Calculate: a) b) c)
Construct a Venn Diagram S A B
0.2
0.4
0.4
Mutually Exclusive, so the circles are separate 5F

45. Construct a Venn Diagram
Probability
Mutually Exclusive and Independent Events
Events C and D are Independent and P(C) = 1/3 and P(D) = 1/5
Calculate: a) b) c)
Construct a Venn Diagram S C D
4/15
1/15
2/15
8/15
P(C) = 1/3 in total so:
P(D) = 1/3 in total so: 5F

46. 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