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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)=611,P(B)=511P(A) = \frac{6}{{11}},P(B) = \frac{5}{{11}}andP(AB)=711P(A \cup B) = \frac{7}{{11}}, find

i.P(AB)P(A \cap B)
ii.P(A/B)P(A/B)
iii.P(B/A)P(B/A)
A.0.58,0.58,0.630.58,0.58,0.63
B.0.42,0.67,0.570.42,0.67,0.57
C.0.36,0.80,0.660.36,0.80,0.66
D.0.80,0.98,0.870.80,0.98,0.87

Explanation

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)=611P(A) = \frac{6}{{11}} it is also given that the
probability of the event B i.e.., P(B)=511P(B) = \frac{5}{{11}} and the probability of the occurrence
of events A or B i.e.., P(AB)=711P(A \cup B) = \frac{7}{{11}}
i. To find P(AB)P(A \cap B) i.e.., the probability of the events AA and BB
As we know the Addition theorem on probability i.e..,
P(AB)=P(A)+P(B)P(AB)P(A \cup B) = P(A) + P(B) - P(A \cap B) (1) \to (1)
Now, let us put the given values of P(A)P(A), P(B)P(B)and P(AB)P(A \cup B) in the equation 1, we get

711=611+511P(AB) P(AB)=611+511711 P(AB)=6+5711 P(AB)=411=0.36  \frac{7}{{11}} = \frac{6}{{11}} + \frac{5}{{11}} - P(A \cap B) \\\ \Rightarrow P(A \cap B) = \frac{6}{{11}} + \frac{5}{{11}} - \frac{7}{{11}} \\\ \Rightarrow P(A \cap B) = \frac{{6 + 5 - 7}}{{11}} \\\ P(A \cap B) = \frac{4}{{11}} = 0.36 \\\
Hence, the probability of the occurrence of both the events A and B is 0.360.36

ii.To find P(A/B)P(A/B)i.e.., the probability of the event A after the occurrence of event B
So, to find the P(A/B)P(A/B) let us consider the concept of conditional probability i.e..,
P(A/B)=P(AB)P(B)(2)P(A/B) = \frac{{P(A \cap B)}}{{P(B)}} \to (2)
Since we have the values of P(AB)P(A \cap B) and P(B)P(B), let us substitute in equation 2, we get
P(A/B)=411511=45=0.80\Rightarrow P(A/B) = \frac{{\frac{4}{{11}}}}{{\frac{5}{{11}}}} = \frac{4}{5} = 0.80
Hence, the required value of P(A/B)P(A/B)is 0.80

iii.To find P(B/A)P(B/A) i.e.., the probability of the event B after the occurrence of event A
So, to find the P(B/A)P(B/A) let us consider the concept of conditional probability i.e..,
P(B/A)=P(AB)P(A)(3)P(B/A) = \frac{{P(A \cap B)}}{{P(A)}} \to (3)
Since we have the values of P(AB)P(A \cap B) andP(A)P(A), let us substitute in equation 3, we get
P(B/A)=411611=46=23=0.66P(B/A) = \frac{{\frac{4}{{11}}}}{{\frac{6}{{11}}}} = \frac{4}{6} = \frac{2}{3} = 0.66
Hence, the required value of P(B/A)P(B/A)is 0.66
Hence the correct option for the given question is ‘C’

Note: Here, in this question P(A/B)P(A/B)and P(B/A)P(B/A) both are considered as the conditional probability where P(A/B)P(A/B)is the “probability of the event A after the occurrence of event B”
and P(B/A)P(B/A) is the “probability of the event B after the occurrence of event A”