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

The correct answer is Option (2) → 

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

$⇒P(X = x_1) =\frac{k}{2}, P(X = x_2) =\frac{k}{3}, P(X = x_3) = k, P(X = x_4) =\frac{k}{5}$.

We know that $ΣP(X = x_i) = 1⇒\frac{k}{2}+\frac{k}{3}+k+\frac{k}{5}=1$

$⇒\left(\frac{1}{2}+\frac{1}{3}+1+\frac{1}{5}\right)k=1⇒\frac{15+10+30+6}{30}k=1$

$⇒\frac{61}{30}k=1⇒k=\frac{30}{61}$

$∴\frac{k}{2}=\frac{15}{61},\frac{k}{3}=\frac{10}{61},\frac{k}{5}=\frac{6}{61}$

∴ The probability distribution of $X=\begin{pmatrix}x_1&x_2& x_3& x_4\\\frac{15}{61}&\frac{10}{61}&\frac{30}{61}&\frac{6}{61}\end{pmatrix}$