The document discusses laws of probability and conditional probability. It provides examples of calculating conditional probability from tables and applying laws such as P(A ∪ B) = P(A) + P(B) - P(A ∩ B) to probability problems. Conditional probability is calculated using tables showing numbers of people by sex and whether they wear glasses to determine probabilities such as P(G|F).

2. 2.2 Laws of probability - Targets
1. Calculate conditional probability from a table.
2. Apply the law P(A ∪ B) = P(A) + P(B) – P(A ∩ B) to all
events.
3. Apply the law P(A ∩ B) = P(A)P(B I A) to events which
are not independent.
4. Solve probability problems using the laws of
probability or tree diagrams.
Stretch and challenge:
5. How does a tree diagram with 2 outcomes on
each branch link in with the binomial
expansion/probability?

3. Conditional probability
• A room contains 25 people. The table below shows
the numbers of each sex and whether or not they
are wearing glasses.
• A person is selected at random.
• F is the event that the person selected is female.
• G is the event that the person selected is wearing
glasses.

4. Conditional probability
• There are a total of 9 people wearing glasses so P(G) =
9
25
.
• Only 4 of the nine males are wearing glasses so the
probability of a male wearing glasses is
4
9
while for the
females the probability is
5
16
.
• This means the probability of event G occurring is affected
by whether or not event F has occurred.
• The 2 events are not independent.

5. Conditional probability
• The conditional probability that the
person selected is wearing glasses given
that they are female is denoted P(G I F).
P(A I B) denotes the probability that event A
happens given that event B happens.
Two events A and B are independent if:
P(A) = P(A I B)

6. Example 1 – 3.5 Page 44/45
Students on the first year of a science course at a university
take an optional language module. The number of students of
each sex choosing each available language is shown below.
A student is selected at random
M is the probability that the student selected is male
R is the event that the student selected is studying Russian
Write down the value of:
a) P(M)
There are 40 male students out of a total of 70 students.
40
70
= 0.571

7. Example 1 – 3.5 Page 44/45
Students on the first year of a science course at a university
take an optional language module. The number of students of
each sex choosing each available language is shown below.
A student is selected at random
M is the probability that the student selected is male
R is the event that the student selected is studying Russian
Write down the value of:
b) P(R)
21 students are studying Russian.
P(R) =
21
70
= 0.3

8. Example 1 – 3.5 Page 44/45
Students on the first year of a science course at a university
take an optional language module. The number of students of
each sex choosing each available language is shown below.
A student is selected at random
M is the probability that the student selected is male
R is the event that the student selected is studying Russian
Write down the value of:
c) P(M I R)
There are 21 students studying Russian and 14 are male.
14
21
= 0.67

9. Example 1 – 3.5 Page 44/45
Students on the first year of a science course at a university
take an optional language module. The number of students of
each sex choosing each available language is shown below.
A student is selected at random
M is the probability that the student selected is male
R is the event that the student selected is studying Russian
Write down the value of:
d) P(M ∩ R)
There are 14 students who are both male and studying Russian
14
70
= 0.2

10. Example 1 – 3.5 Page 44/45
Students on the first year of a science course at a university
take an optional language module. The number of students of
each sex choosing each available language is shown below.
A student is selected at random
M is the probability that the student selected is male
R is the event that the student selected is studying Russian
Write down the value of:
e) P(M ∪ R)
There are 17 + 9 + 14 + 7 = 47 students who are either male or
studying Russian (or both).
47
70
= 0.671

11. Example 1 – 3.5 Page 44/45
Students on the first year of a science course at a university
take an optional language module. The number of students of
each sex choosing each available language is shown below.
A student is selected at random
M is the probability that the student selected is male
R is the event that the student selected is studying Russian
Write down the value of:
f) P(R I M)
There are 40 male students of which 14 are studying Russian
14
40
= 0.35

12. Example 1 – 3.5 Page 44/45
Students on the first year of a science course at a university
take an optional language module. The number of students of
each sex choosing each available language is shown below.
A student is selected at random
M is the probability that the student selected is male
R is the event that the student selected is studying Russian
Write down the value of:
g) P(M’)
There are 30 students who are not male
𝟑𝟎
𝟕𝟎
= 0.429

13. Example 1 – 3.5 Page 44/45
Students on the first year of a science course at a university
take an optional language module. The number of students of
each sex choosing each available language is shown below.
A student is selected at random
M is the probability that the student selected is male
R is the event that the student selected is studying Russian
Write down the value of:
h) P(R’)
There are 29 + 20 = 49 students not studying Russian
49
70
= 0.7

14. Example 1 – 3.5 Page 44/45
Students on the first year of a science course at a university
take an optional language module. The number of students of
each sex choosing each available language is shown below.
A student is selected at random
M is the probability that the student selected is male
R is the event that the student selected is studying Russian
Write down the value of:
i) P(R I M’)
Of the 30 who are not male, 7 are studying Russian.
7
30
= 0.233

15. Example 1 – 3.5 Page 44/45
Students on the first year of a science course at a university
take an optional language module. The number of students of
each sex choosing each available language is shown below.
A student is selected at random
M is the probability that the student selected is male
R is the event that the student selected is studying Russian
Write down the value of:
j) P(R’ I M)
Of the 40 males, 17 + 9 = 26 are not studying Russian
26
40
= 0.65

16. Example 1 – 3.5 Page 44/45
Students on the first year of a science course at a university
take an optional language module. The number of students of
each sex choosing each available language is shown below.
A student is selected at random
M is the probability that the student selected is male
R is the event that the student selected is studying Russian
Write down the value of:
k) P(M’ ∩ R)
There are 7 students who are not male and are studying
Russian,
7
70
= 0.1

17. Example 1 – 3.5 Page 44/45
Students on the first year of a science course at a university
take an optional language module. The number of students of
each sex choosing each available language is shown below.
A student is selected at random
M is the probability that the student selected is male
R is the event that the student selected is studying Russian
Write down the value of:
l) P(M ∪ R’)
There are 40 + 12 + 11 = 63 students who are either male or not
studying Russian (or both),
63
70
= 0.9.

18. Addition law of probability
• In 2.1 we saw that if A and B are mutually
exclusive events then P(A ∪ B) = P(A) + P(B).
• A more general form of this expression which
applies whether or not A and B are mutually
exclusive events is
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
This is known as the addition law of
probability

19. Addition law of probability
• Diagram on page 45 and subsequent
working on 45/46 illustrates this
• Also see next example for how a numerical
example proof

20. Example 2
150 students at a catering college have to
choose between 3 tasks as part of their final
assessment. The table on example 3.6 on page
46 gives their choices.
A student is randomly selected.
C denotes the event that the selected student
chooses cake baking.
D denotes the event that the selected student is
female.
Verify that P(C ∪ D) = P(C) + P(D) – P(C ∩ D)

21. Example 2
P(C ∪ D) = P(C) + P(D) – P(C ∩ D)
P(C) => There are 54 + 26 = 80 cake-baking students.
P(C) =
80
150
P(D) => There are 26 + 18 + 16 = 60 female students
P(D) =
60
150
P(C ∩ D) => There are 26 female cake-baking students.
P(C ∩ D) =
26
150
P(C) + P(D) – P(C ∩ D) =
80
150
+
60
150
-
26
150
P(C) + P(D) – P(C ∩ D) =
114
150
There are 54 + 26 + 18 + 16 = 114 students who are either cake-
baking or female or both
P(C ∪ D) =
114
150
= P(C) + P(D) – P(C ∩ D) as required.

22. Task 1
• Exercise 3E
• Questions 1 and 3.

23. Task 1 Answers1a
5
8
1b
2
3
1c
3
8
1d
47
120
1e
9
10
1f
47
80
1g
28
75
1h
7
10
1i
33
80
1j
7
2a
23
30
2b
121
150
2c
25
29
2d
119
150
2e
1
6
2f
3
5
2g
2
5
2h
90
121
2i
31
35
3a
4
9
3b
2
3
3c
5
9
3d
2
15
3e
4
5
3f
13
23
3g
1
5
3h
7
10
3i 0
3j A and C not
independent

24. Multiplication law
• If A and B are independent events, then
P(A ∩ B) = P(A)P(B).
• This is a special case of the multiplication
law for probability.
P(A ∩ B) = P(A)P(B I A)

25. Multiplication law
• Verify this by example
• A person is selected at random
• F is the event that the person is selected is female.
• G is the event that the person selected is wearing glasses.
• P(F ∩ G), the probability that the person selected is a female
wearing glasses is
5
25
or 0.2.
• P(F) =
16
25
= 0.64 and P(G I F) =
5
16
= 0.3125
• P(F)P(G I F) = 0.2
• P(F)P(G I F) = P(F ∩ G)

26. Example 3 – 3.7 Page 49
Sheena buys ten apparently identical oranges.
Unknown to her, two of the oranges are rotten.
She selects two of the ten oranges at random
and gives them to her grandson.
Find the probability that:
a) Both the oranges are rotten
b) Exactly one of the oranges is rotten.

27. Example 4 – 3.8 Page 50
When Bali is on holiday she intends to go for a five-
mile run before breakfast each day. However,
sometimes she stays in bed instead.
The probability that she will go for a run on the first
morning is 0.7.
Thereafter, the probability she will go for a run is 0.7
IF she went on the previous morning and 0.6 if she
did not.
Find the probability that on the first 3 days of the
holiday she will go for:
a) 3 runs
b) Exactly 2 runs

28. Task 2
• Exercise 3F
• Pages 50-52
• Questions 1, 3, 5 and 6

29. Task 2 Answers
1a 0.195
1b 0.499
3ai 0.0152
3aii 0.182
3aiii 0.227
3bi 0.0909
3bii 0.136
3biii 0.409
3biv 0.218
5a 0.196
5b 0.0240
5c 0.0840
5d 0.0960
5e 0.240
5f 0.228
5g 0.192
6a 0.0461
6b 0.233
6c 0.0121
6d 0.279
6e 0.0101
6f 0.266
6g 0.152

30. 2.2 Laws of probability – Key Points
• P(A I B) denotes the probability that event A
happens given that event B happens.
• Two events A and B are independent if:
P(A) = P(A I B)
• Addition law of probability
– P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
• Multiplication law of probability
– P(A ∩ B) = P(A)P(B I A)