Let X be random variable which assumes $x_1, x_2, x_3, x_4$ such that $2P(X=x_1)= 3P(X=x_2)=P(X=x_3)=5P(X=x_4)$, then the probability distribution of X is

The correct answer is Option (4) ** 

Given: $2P(X=x_1)=3P(X=x_2)=P(X=x_3)=5P(X=x_4)$

Let the common value be $k$.

$2P(X=x_1)=k \Rightarrow P(X=x_1)=\frac{k}{2}$

$3P(X=x_2)=k \Rightarrow P(X=x_2)=\frac{k}{3}$

$P(X=x_3)=k$

$5P(X=x_4)=k \Rightarrow P(X=x_4)=\frac{k}{5}$

Total probability = $1$

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

$k\left(\frac{15+10+30+6}{30}\right)=1$

$k\left(\frac{61}{30}\right)=1$

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

Thus,

$P(X=x_1)=\frac{k}{2}=\frac{15}{61}$

$P(X=x_2)=\frac{k}{3}=\frac{10}{61}$

$P(X=x_3)=k=\frac{30}{61}$

$P(X=x_4)=\frac{k}{5}=\frac{6}{61}$

The probability distribution is: $(\frac{15}{61},\frac{10}{61},\frac{30}{61},\frac{6}{61})$.