Question
Question: If \(P(A) = \frac{6}{{11}},P(B) = \frac{5}{{11}}\)and\(P(A \cup B) = \frac{7}{{11}}\), find i.\...
If P(A)=116,P(B)=115andP(A∪B)=117, find
i.P(A∩B)
ii.P(A/B)
iii.P(B/A)
A.0.58,0.58,0.63
B.0.42,0.67,0.57
C.0.36,0.80,0.66
D.0.80,0.98,0.87
Solution
Hint: Here, to solve the given problem we use the Addition theorem of probability
as well as the conditional probability Concepts.
Given, the Probability of the event A i.e.., P(A)=116 it is also given that the
probability of the event B i.e.., P(B)=115 and the probability of the occurrence
of events A or B i.e.., P(A∪B)=117
i. To find P(A∩B) i.e.., the probability of the events A and B
As we know the Addition theorem on probability i.e..,
P(A∪B)=P(A)+P(B)−P(A∩B) →(1)
Now, let us put the given values of P(A), P(B)and P(A∪B) in the equation 1, we get
117=116+115−P(A∩B) ⇒P(A∩B)=116+115−117 ⇒P(A∩B)=116+5−7 P(A∩B)=114=0.36
Hence, the probability of the occurrence of both the events A and B is 0.36
ii.To find P(A/B)i.e.., the probability of the event A after the occurrence of event B
So, to find the P(A/B) let us consider the concept of conditional probability i.e..,
P(A/B)=P(B)P(A∩B)→(2)
Since we have the values of P(A∩B) and P(B), let us substitute in equation 2, we get
⇒P(A/B)=115114=54=0.80
Hence, the required value of P(A/B)is 0.80
iii.To find P(B/A) i.e.., the probability of the event B after the occurrence of event A
So, to find the P(B/A) let us consider the concept of conditional probability i.e..,
P(B/A)=P(A)P(A∩B)→(3)
Since we have the values of P(A∩B) andP(A), let us substitute in equation 3, we get
P(B/A)=116114=64=32=0.66
Hence, the required value of P(B/A)is 0.66
Hence the correct option for the given question is ‘C’
Note: Here, in this question P(A/B)and P(B/A) both are considered as the conditional probability where P(A/B)is the “probability of the event A after the occurrence of event B”
and P(B/A) is the “probability of the event B after the occurrence of event A”