Question

If $A$ and $B$ are two independent events such that $P\left(A\right)=\frac{1}{2}$ and $P\left(B\right)=\frac{1}{3}$, then $P$(neither $A$ nor $B$) is equal to

• A. $\frac{2}{3}$
• B. $\frac{1}{6}$
• C. $\frac{5}{6}$
• D. $\frac{1}{3}$

Answer 1

The correct option is D

$\frac{1}{3}$

Explanation for the correct option:

Step 1. Find the value of $P$(neither $A$ nor $B$):

Given that, $P\left(A\right)=\frac{1}{2}$ and $P\left(B\right)=\frac{1}{3}$

$P$(neither $A$ nor $B$) $=P\left(A\text{'}\cap B\text{'}\right)$

$$\begin{gathered} =1-P\left(A\cup B\right) \\ =1-\left[P\left(A\right)+P\left(B\right)-P(A\cap B)\right] \end{gathered}$$

$\because A$ and $B$ are independent events,

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

$\Rightarrow$$P(A\text{'}\cap B\text{'})=1-\left[P\right(A)+P(B)-P(A).P(B\left)\right]$

Step 2. Substituting the values of $P\left(A\right)$ and $P\left(B\right)$:

$\begin{array}{rcl}P(A\text{'}\cap B\text{'})& =& 1-\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{2}\times \frac{1}{3}\right)\\ & =& 1-\left(\frac{5}{6}-\frac{1}{6}\right)\\ & =& 1-\frac{4}{6}\\ & =& \frac{2}{6}\\ & =& \frac{\mathbf{1}}{\mathbf{3}}\end{array}$

Hence, Option ‘D’ is Correct.