Question

If \[P(A)=P(B)=x\] and \[P(A\cap B)=P(\overline{A}\cap \overline{B})=\dfrac{1}{3}\], then \[x=\]
A. \[\dfrac{1}{2}\]
B. \[\dfrac{1}{3}\]
C. \[\dfrac{1}{4}\]
D. \[\dfrac{1}{6}\]

Answer 1

Hint: In the above question, we are to find the value of the variable $x$ which represents the value of the probabilities of the events $A$ and $B$. In order to get the value of the variables, we should know the formula which represents the relation between the union and the intersection of two events.



Formula Used:A probability is the ratio of favourable outcomes of an event to the total number of outcomes. So, the probability lies 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, the complement of the union/intersection of two sets is equal to the intersection/union of complemented sets respectively. I.e.,
$\begin{align}
  & (\overline{A\cup B})={{A}^{'}}\cap {{B}^{'}} \\
 & (\overline{A\cap B})={{A}^{'}}\cup {{B}^{'}} \\
\end{align}$



Complete step by step solution:Here, we are given the values of the probability of common events between $A$ and $B$, i.e.,
\[P(A\cap B)=\dfrac{1}{3}\]
And the probability of non-occurrence of the event and , i.e.,
\[P(\overline{A}\cap \overline{B})=\dfrac{1}{3}\]
We can write it as,
\[\begin{align}
  & P(\overline{A}\cap \overline{B})=1-P(A\cup B) \\
 & P(A\cup B)=1-P(\overline{A}\cap \overline{B}) \\
 & \text{ }=1-\dfrac{1}{3} \\
 & \text{ }=\dfrac{2}{3} \\
\end{align}\]
By the addition theorem on probability,
\[P(A\cup B)=P(A)+P(B)-P(A\cap B)\]
On substituting the given values,
\[\begin{align}
  & \text{ }\dfrac{2}{3}=x+x-\dfrac{1}{3} \\
 & \Rightarrow 2x=\dfrac{2}{3}+\dfrac{1}{3} \\
 & \Rightarrow 2x=1 \\
 & \text{ }\therefore x=\dfrac{1}{2} \\
\end{align}\]



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.