P(A) = 0.4 P(B) = 0.75 P(A B) = 0.35 a)Draw a venn diagram to show this information (3) b)Calculate...
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P(A) = 0.4 P(B) = 0.75 P(A B) = 0.35 a) Draw a venn diagram to show this information (3) b) Calculate P(A B) (1) c) Calculate P(A` B`) (1) d) Calculate P(A` B) (1)
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Transcript of P(A) = 0.4 P(B) = 0.75 P(A B) = 0.35 a)Draw a venn diagram to show this information (3) b)Calculate...
P(A) = 0.4
P(B) = 0.75
P(A B) = 0.35
a) Draw a venn diagram to show this information (3)
b) Calculate P(A B) (1)
c) Calculate P(A` B`) (1)
d) Calculate P(A` B) (1)
A B
0.350.05 0.4
S
0.2
a) P(A B) = 0.4 + 0.75 – 0.35 = 0.8b) P(A` B`) = 0.2c) P(A` B) = 0.95
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
CONDITIONAL PROBABILITY
The probability of B may be different if you know that A has already occurred.
A B
ia - i b - i
S
The probability of B given A = i
a
The probability of P(BA) = P(AB)
P(A)
CONDITIONAL PROBABILITY
In a class of 20 students 10 study French, 9 study Maths and 3 study both
French Maths
37 6
S
The probability they study Maths given that they study French
The probability of P(MF) = P(MF) = 3
P(F) 10
4
P(BA) = P(AB)
P(A)
MULTIPLICATION RULE
P(AB) = P(BA) x P(A)
P(AB) = P(BA)P(A)
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(CD) = P(CD) x P(D)
P(CD) = 0.3 x 0.6 = 0.18
C D
0.180.02 0.42
S
0.38
P(D C) = P(CD) = 0.18 = 0.9
P( C ) 0.2
C D
0.180.02 0.42
S
0.38
P(C`D`) = 0.38
P(C`D) = 0.42
Example
A and B are two events
P(A|B) = 0.1, P(A|B`) = 0.6, P(B) = 0.3
Find
a) P(AB)
b) P(AB`)
c) P(A)
d) P(B|A)
e) P(B|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(AB) b) P(AB`) c) P(A) d) P(B|A) e) P(B|A`)
a)P(AB)=P(A|B)P(B)=0.1x0.3=0.03
b) P(AB`)=P(A|B`)P(B`)=0.6x0.7=0.42
A B
0.030.42 0.27
S
0.28
c)P(A) = 0.42+0.03 = 0.45
d)P(B|A)= 0.03 = 0.06
0.45
e)P(B|A`)= 0.27 = 0.490
0.55
4 a 0.7 b 0.667 c 0.8 d 0.45 a 0.5 b 0.3 c 0.36 a 0.3 b 0.35 c 0.4
7 a 0.0833b 0.15c 0.233d 0.357e 0.643f 0.783
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
1 2 3 4 5
2 3 4 5 6
3 4 5 6 7
4 5 6 7 8
+ 1 2 3 4
1 2 3 4 5
2 3 4 5 6
3 4 5 6 7
4 5 6 7 8
+ 1 2 3 4
1 2 3 4 5
2 3 4 5 6
3 4 5 6 7
4 5 6 7 8
P(A) = at least one 3
P(B) = sum of exactly 5
P(BA) = 2
7
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
1 2 3 4 5
2 3 4 5 6
3 4 5 6 7
4 5 6 7 8
P(A) = at least one 3 = 7/16
P(A B) = 2/16
P(BA) = P(A B) = 2/16
P(A) 7/16
P(BA) = 2/16 ÷ 7/16 = 2/16 x 16/7 = 2/7
Tree diagrams and conditional probability
Event A Event B
P(A)
P(A`)
P(B|A)
P(B`|A)
P(B|A`)
P(B`|A`)
P(AB)=P(A) P(B|A)
P(AB`)=P(A) P(B`|A)
P(A`B)=P(A`) P(B|A`)
P(A`B`)=P(A`) P(B`|A`)
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 0.9. The weather forecast gives a probability of 0.75 that it will rain on the day of the race.
a) Find the probability that there is a big turnout and it rains
b) Find the probability that there is a big turnout
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 0.9. The weather forecast gives a probability of 0.75 that it will rain on the day of the race.
a) Find the probability that there is a big turnout and it rains
b) Find the probability that there is a big turnout
P(R)
P(R`)
P(B)
P(B`)P(B)
P(B`)
0.75
0.25
0.4
0.6
0.9
0.1
a) 0.75 x 0.4 = 0.3
b) 0.3 + (0.25x0.9) = 0.525
Example 2 A and B are two events P(A|B) = 0.1, P(A|B`) = 0.6, P(B) = 0.3
Find a)P(AB) b) P(AB`) c) P(A) d) P(B|A) e) P(B|A`)
P(B)
P(B`)
P(A)
P(A`)P(A)
P(A`)
0.3
0.7
0.1
0.9
0.6
0.4
a)P(AB) = 0.3 x 0.1 = 0.03
b)P(AB`) = 0.7 x 0.6 = 0.42
c)P(A) = P(AB) + P(AB`) = 0.45
d)P(B|A) = P(BA) = 0.03 = 0.06 Note this is exactly the same as the venn diagrams
P(A) 0.45
