Answer
454.5k+ views
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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