Question

If two events are $A$ and $B$ such that $P(A\cup B)=\dfrac{5}{6}, P(A\cap B)=\dfrac{1}{3}$, $P(\overline{B})=\dfrac{1}{3}$ then $P(A)=?$
A. $\dfrac{1}{4}$
B. $\dfrac{1}{3}$
C. $\dfrac{1}{2}$
D. $\dfrac{2}{3}$

Answer 1

Hint: In this question, we are to find the probability of an event. For this question, the addition theorem on probability is used. All the given values are substituted in the addition theorem of probability to find the required probability.

Formula used: The probability is calculated by,
$P(E)=\dfrac{n(E)}{n(S)}$
If there are two events in a sample space, then the addition theorem on probability is given by
$P(A\cup B)=P(A)+P(B)-P(A\cap B)$
When two events happen independently, the occurrence of one is not impacted by the occurrence of the other.
For the events $A$ and $B$, $P(A\cap B)=P(A)\cdot P(B)$ if they are independent and $P(A\cap B)=\Phi $ if they are mutually exclusive.

Complete step by step solution: Consider two events $A$ and \[B\].
It is given that,
\[P(A\cup B)=\dfrac{5}{6}\]
\[P(A\cap B)=\dfrac{1}{3}\]
$P(\overline{B})=\dfrac{1}{3}$
So,
$\begin{align}
  & P(B)=1-P(\overline{B}) \\
 & \text{ }=1-\dfrac{1}{3} \\
 & \text{ }=\dfrac{2}{3} \\
\end{align}$
Then, by substituting in the formula, we get
\[\begin{align}
  & P(A\cup B)=P(A)+P(B)-P(A\cap B) \\
 & \text{ }\dfrac{5}{6}=P(A)+\dfrac{2}{3}-\dfrac{1}{3} \\
 & \text{ }\Rightarrow P(A)=\dfrac{5}{6}-\dfrac{1}{3}=\dfrac{1}{2} \\
\end{align}\]

Thus, Option (C) 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. The actual probability must be calculated.