If $A$ and $B$ are independent events of a random experiment, such that $P(A\cap B)=\frac{1}{6}$, $P(A\text{'}\cap B\text{'})=\frac{1}{3}$, then $P\left(A\right)$ is equal to

The correct option is B

$\frac{1}{3}$

Explanation for the correct option:

Step 1: Find the value of $P(A\cup B)$

we have given $A$ and $B$ are independent events,

$P(A\cap B)=\frac{1}{6}$,$P(A\text{'}\cap B\text{'})=\frac{1}{3}$

we know that, $\begin{array}{rcl}\mathbf{P}\mathbf{(}\mathbf{A}\mathbf{\text{'}}\mathbf{\cap }\mathbf{B}\mathbf{\text{'}}\mathbf{)}& \mathbf{=}& \mathit{P}{\mathbf{(}\mathbf{A}\mathbf{\cup }\mathbf{B}\mathbf{)}}^{\mathbf{\text{'}}}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}P{(A\cup B)}^{\text{'}}& =& 1-P(A\cup B)\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\frac{1}{3}& =& 1-P(A\cup B)\end{array}$

$\Rightarrow$ $\begin{array}{rcl}P(A\cup B)& =& 1-\frac{1}{3}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}P(A\cup B)& =& \frac{2}{3}\end{array}$

Step 2: Find the value of$\mathbf{}P\left(A\right)$.

We know, $\begin{array}{rcl}\mathit{P}\mathbf{(}\mathbf{A}\mathbf{\cup }\mathbf{B}\mathbf{)}& \mathbf{=}& \mathbf{P}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{+}\mathbf{P}\mathbf{\left(}\mathbf{B}\mathbf{\right)}\mathbf{-}\mathbf{P}\mathbf{(}\mathbf{A}\mathbf{\cap }\mathbf{B}\mathbf{)}\end{array}$

$\begin{array}{rcl} P(A\cup B)& =& \frac{2}{3},P(A\cap B)=\frac{1}{6}\\ \therefore P(A\cup B)+P(A\cap B)& =& P\left(A\right)+P\left(B\right)\end{array}$

$\Rightarrow$ $\begin{array}{rcl}P\left(A\right)+P\left(B\right)& =& \frac{2}{3}+\frac{1}{6}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}P\left(A\right)+P\left(B\right)& =& \frac{5}{6}.....\left(1\right)\end{array}$

we have given that $A$ and$B$ are independent events,

$\therefore P(A\cap B)=P\left(A\right).P\left(B\right)......\left(2\right)$

From equation$\left(1\right)$ and $\left(2\right)$, we get

$\mathit{P}\mathbf{\left(}\mathit{A}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}$

Hence, the option$\left(B\right)$ is correct.