Answer

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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.., $P(A) = \frac{6}{{11}}$ it is also given that the

probability of the event B i.e.., $P(B) = \frac{5}{{11}}$ and the probability of the occurrence

of events A or B i.e.., $P(A \cup B) = \frac{7}{{11}}$

i. To find $P(A \cap B)$ i.e.., the probability of the events $A$ and $B$

As we know the Addition theorem on probability i.e..,

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ \to (1)$

Now, let us put the given values of $P(A)$, $P(B)$and $P(A \cup B)$ in the equation 1, we get

$

\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 $0.36$

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..,

$P(A/B) = \frac{{P(A \cap B)}}{{P(B)}} \to (2)$

Since we have the values of $P(A \cap B)$ and $P(B)$, let us substitute in equation 2, we get

$ \Rightarrow P(A/B) = \frac{{\frac{4}{{11}}}}{{\frac{5}{{11}}}} = \frac{4}{5} = 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..,

$P(B/A) = \frac{{P(A \cap B)}}{{P(A)}} \to (3)$

Since we have the values of $P(A \cap B)$ and$P(A)$, let us substitute in equation 3, we get

$P(B/A) = \frac{{\frac{4}{{11}}}}{{\frac{6}{{11}}}} = \frac{4}{6} = \frac{2}{3} = 0.66$

Hence, the required value of $P(B/A)$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”

as well as the conditional probability Concepts.

Given, the Probability of the event A i.e.., $P(A) = \frac{6}{{11}}$ it is also given that the

probability of the event B i.e.., $P(B) = \frac{5}{{11}}$ and the probability of the occurrence

of events A or B i.e.., $P(A \cup B) = \frac{7}{{11}}$

i. To find $P(A \cap B)$ i.e.., the probability of the events $A$ and $B$

As we know the Addition theorem on probability i.e..,

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ \to (1)$

Now, let us put the given values of $P(A)$, $P(B)$and $P(A \cup B)$ in the equation 1, we get

$

\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 $0.36$

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..,

$P(A/B) = \frac{{P(A \cap B)}}{{P(B)}} \to (2)$

Since we have the values of $P(A \cap B)$ and $P(B)$, let us substitute in equation 2, we get

$ \Rightarrow P(A/B) = \frac{{\frac{4}{{11}}}}{{\frac{5}{{11}}}} = \frac{4}{5} = 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..,

$P(B/A) = \frac{{P(A \cap B)}}{{P(A)}} \to (3)$

Since we have the values of $P(A \cap B)$ and$P(A)$, let us substitute in equation 3, we get

$P(B/A) = \frac{{\frac{4}{{11}}}}{{\frac{6}{{11}}}} = \frac{4}{6} = \frac{2}{3} = 0.66$

Hence, the required value of $P(B/A)$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”

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