Question

Choose the correct alternative for each of the following. If $P\left(E_1\right)=\frac{1}{6}$, $P\left(E_2\right)=\frac{1}{3}$, $P\left(E_3\right)=\frac{1}{6}$, where $E_1$, $E_2$, $E_3$, $E_4$ are elementary events of a random experiment, then P($E_4$) is equal to

• A. $\frac{1}{2}$
• B. $\frac{2}{3}$
• C. $\frac{1}{3}$
• D. None

Answer 1

The correct option is C $\frac{1}{3}$

Given, probabilities of different elements of the event:
$P\left(E_1\right)=\frac{1}{6}$, $P\left(E_2\right)=\frac{1}{3}$, $P\left(E_3\right)=\frac{1}{6}$
The sum of individual probabilities $=1$
$P\left(E_1\right)+P\left(E_2\right)+P\left(E_3\right)+P\left(E_4\right)=1$
$\frac{1}{6}+\frac{1}{3}+\frac{1}{6}+P\left(E_4\right)=1$
$\frac{2}{3}+P\left(E_4\right)=1$
$P\left(E_4\right)=\frac{1}{3}$