Let X be a random variable which assumes value $x_1, x_2, x_3, x_4$ such that $2P(x=x_1)=3P(x=x_2)=P(x=x_3)= 5P(x=x_4), $ then Probability $P(x=x_3) $ will be

$\frac{30}{61}$

The correct answer is Option (2) → $\frac{30}{61}$

The sum of probabilities is,

$\frac{k}{2}+\frac{1}{3}k+k+\frac{k}{5}=1$

$\frac{61k}{30}=1$

$k=\frac{30}{61}$