Question

If \[A\] and \[B\] are events such that \[P(A\cup B)=\dfrac{3}{4}\], \[P(A\cap B)=\dfrac{1}{4}\], and \[P(\overline{A})=\dfrac{2}{3}\], then \[P(\overline{A}\cap B)\] is
A. \[\dfrac{5}{12}\]
B. \[\dfrac{3}{8}\]
C. \[\dfrac{5}{8}\]
D. \[\dfrac{1}{4}\]

Answer 1

Hint: In the above question, we are asked to find the value of the expression \[P(\overline{A}\cap B)\]. In order to get the value of the given expression, the addition theorem on probability is used. The identities from set theory are used.



Formula Used:A probability is the ratio of favourable outcomes of an event to the total number of outcomes. So, the probability lies in between 0 and 1.
The probability is calculated by,
\[P(E)=\dfrac{n(E)}{n(S)}\]
\[n(E)\] - favourable outcomes and \[n(S)\] - sample.
If there are two events in a sample space, then the addition theorem on probability is given by
\[P(A\cup B)=P(A)+P(B)-P(A\cap B)\]
According to set theory,
Union of sets is described as:
\[A\cup B=\{x/x\in A\text{ or }x\in B\}\]
Intersection of sets is described as:
\[A\cap B=\{x/x\in A\text{ and }x\in B\}\]
Symmetric difference is described as:
\[\begin{align}
  & A-B=\{x/x\in A\text{ and }x\notin B\} \\
 & B-A=\{x/x\in B\text{ and }x\notin A\} \\
\end{align}\]



Complete step by step solution:Here, we are given the values for the probability of occurrence of both \[A\] and \[B\] events which is equal to \[\dfrac{3}{4}\], the probability of not occurrence of any of the events is equal to \[\dfrac{1}{4}\], and the probability of non-occurrence of the event \[A\] is \[\dfrac{2}{3}\].

According to the formula of addition theorem on probability of the two events \[A\] and \[B\],
\[P(A\cup B)=P(A)+P(B)-P(A\cap B)\text{ }...(1)\]
But we have \[P(\overline{A})=\dfrac{2}{3}\], then
\[\begin{align}
  & P(A)=1-P(\overline{A}) \\
 & \text{ }=1-\dfrac{2}{3} \\
 & \text{ }=\dfrac{1}{3} \\
\end{align}\]
Therefore, we can put all the known values in (1), and we get-
\[\begin{align}
  & \dfrac{3}{4}=\dfrac{1}{3}+P(B)-\dfrac{1}{4} \\
 & \Rightarrow P(B)=\dfrac{3}{4}-\dfrac{1}{3}+\dfrac{1}{4} \\
 & \Rightarrow P(B)=\dfrac{9-4+3}{12} \\
 & \Rightarrow P(B)=\dfrac{8}{12}=\dfrac{2}{3} \\
\end{align}\]
Then, the required probability is calculated by,
\[P(\overline{A}\cap B)=P(B)-P(A\cap B)\]
On substituting,
$\begin{align}
  & P(\overline{A}\cap B)=\dfrac{2}{3}-\dfrac{1}{4} \\
 & \text{ =}\dfrac{8-3}{12} \\
 & \text{ }=\dfrac{5}{12} \\
\end{align}$



Option ‘A’ is correct

Note: In this question, the addition theorem on probability is applied for finding the required probability. By substituting the appropriate values, the required probability is calculated. Here we may go wrong with the complimented probability.