1. P(A) = 0.4
P(B) = 0.75
P(A  B) = 0.35
Draw a venn diagram to show this information (3)
Calculate P(A  B) (1)
Calculate P(A`  B`) (1)
Calculate P(A`  B) (1)

2. A B
0.35
0.05
0.4 S
0.2
P(A  B) = – 0.35 = 0.8
P(A`  B`) = 0.2
P(A`  B) = 0.95

3. CONDITIONAL PROBABILITY
To understand conditional probability
To understand and use the addition, multiplication and conditional probability rules
To be able to use tree diagrams for conditional probability

4. CONDITIONAL PROBABILITY
The probability of B may be different if you know that A has already occurred. A B S
a - i i
b - i
The probability of B given A = i a
The probability of P(BA) = P(AB)
P(A)

5. CONDITIONAL PROBABILITY
In a class of 20 students 10 study French, 9 study Maths and 3 study both
French
Maths S 7 3 6 4
The probability they study Maths given that they study French
The probability of P(MF) = P(MF) = 3
P(F)

6. MULTIPLICATION RULE
P(BA) = P(AB)
P(A)
P(AB) = P(BA) x P(A)
P(AB) = P(BA)P(A)

7. EXAMPLE
P(C) = 0.2
P(D) = 0.6
P(C D) = 0.3
Calculate
a) P(D  C)
P(C`D`)
P (C`D)
P(CD) = P(CD) x P(D)
P(CD) = 0.3 x 0.6 = 0.18 C D S
0.02
0.18
0.42
0.38

8. C D
0.18
0.02
0.42 S
0.38
P(D  C) = P(CD) = = 0.9
P( C )
P(C`D`) = 0.38
P(C`D) = 0.42

9. P(A|B) = 0.1, P(A|B`) = 0.6, P(B) = 0.3 Find
Example
A and B are two events
P(A|B) = 0.1, P(A|B`) = 0.6, P(B) = 0.3
Find
P(AB)
P(AB`)
P(A)
P(B|A)
P(B|A`)

10. Example A and B are two events P(A|B) = 0.1, P(A|B`) = 0.6, P(B) = 0.3
Find a)P(AB) b) P(AB`) c) P(A) d) P(B|A) e) P(B|A`)
a)P(AB)=P(A|B)P(B)=0.1x0.3=0.03
b) P(AB`)=P(A|B`)P(B`)=0.6x0.7=0.42 A B S
0.42
0.03
0.27
0.28
c)P(A) = = 0.45
d)P(B|A)= 0.03 = 0.06
0.45
e)P(B|A`)= 0.27 = 0.490
0.55

12. 4 a 0.7
b 0.667
c 0.8
d 0.4
5 a 0.5
b 0.3
c 0.3
6 a 0.3
b 0.35
c 0.4
7 a
b 0.15
c 0.233
d 0.357
e 0.643
f 0.783

13. EXAMPLE 1
2 fair spinners are numbered 1 to 4. They are spun and the sum of the numbers are recorded.
Given that at least one spinner lands on a 3, find the probability that the spinners sum exactly 5. + 1 2 3 4 5 6 7 8 + 1 2 3 4 5 6 7 8 + 1 2 3 4 5 6 7 8
P(A) = at least one 3
P(B) = sum of exactly 5
P(BA) = 2 7

14. EXAMPLE 1
2 fair spinners are numbered 1 to 4. They are spun and the sum of the numbers are recorded.
Given that at least one spinner lands on a 3, find the probability that the spinners sum exactly 5. + 1 2 3 4 5 6 7 8
P(A) = at least one 3 = 7/16
P(A  B) = 2/16
P(BA) = P(A  B) = 2/16
P(A) /16
P(BA) = 2/16 ÷ 7/16 = 2/16 x 16/7 = 2/7

15. Tree diagrams and conditional probability
P(A)
P(A`)
P(B|A)
P(B`|A)
P(B|A`)
P(AB)=P(A) P(B|A)
P(AB`)=P(A) P(B`|A)
P(A`B)=P(A`) P(B|A`)
P(B`|A`)
P(A`B`)=P(A`) P(B`|A`)
Event A
Event B

16. Example 1
The turnout at an event is dependent on the weather. On a rainy day the probability of a big turnout is 0.4, but if it does not rain, the probability of a big turnout increases to The weather forecast gives a probability of 0.75 that it will rain on the day of the race.
Find the probability that there is a big turnout and it rains
Find the probability that there is a big turnout

17. Example 1
The turnout at an event is dependent on the weather. On a rainy day the probability of a big turnout is 0.4, but if it does not rain, the probability of a big turnout increases to The weather forecast gives a probability of 0.75 that it will rain on the day of the race.
Find the probability that there is a big turnout and it rains
Find the probability that there is a big turnout
a) 0.75 x 0.4 = 0.3
P(B)
0.4
P(R)
b) (0.25x0.9) = 0.525
0.75
0.6
P(B`)
P(B)
0.25
0.9
P(R`)
0.1
P(B`)

18. P(A) 0.1 P(B) 0.3 0.9 P(A`) P(A) 0.6 0.7 P(B`) 0.4 P(A`)
Example A and B are two events P(A|B) = 0.1, P(A|B`) = 0.6, P(B) = 0.3
Find a)P(AB) b) P(AB`) c) P(A) d) P(B|A) e) P(B|A`)
P(A)
0.1
P(B)
0.3
0.9
P(A`)
P(A)
0.7
0.6
P(B`)
0.4
P(A`)
a)P(AB) = 0.3 x 0.1 = 0.03
b)P(AB`) = 0.7 x 0.6 = 0.42
c)P(A) = P(AB) + P(AB`) = 0.45
d)P(B|A) = P(BA) = 0.03 = Note this is exactly the same as the venn diagrams
P(A)