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

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

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

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Expressing each probability in terms of $k$:

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

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

$P(X=x_3) = k$

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

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Since all probabilities must sum to 1:

$P(X=x_1) + P(X=x_2) + P(X=x_3) + P(X=x_4) = 1$

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

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Finding common denominator (LCM of 2, 3, 1, 5 is 30):

$\dfrac{15k}{30} + \dfrac{10k}{30} + \dfrac{30k}{30} + \dfrac{6k}{30} = 1$

$\dfrac{15k + 10k + 30k + 6k}{30} = 1$

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

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

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Substituting $k = \dfrac{30}{61}$:

$P(X=x_1) = \dfrac{k}{2} = \dfrac{30}{61} \times \dfrac{1}{2} = \dfrac{15}{61}$

$P(X=x_2) = \dfrac{k}{3} = \dfrac{30}{61} \times \dfrac{1}{3} = \dfrac{10}{61}$

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

$P(X=x_4) = \dfrac{k}{5} = \dfrac{30}{61} \times \dfrac{1}{5} = \dfrac{6}{61}$

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The probability distribution of $X$ is:

$X$	$x_1$	$x_2$	$x_3$	$x_4$
$P(X)$	$\dfrac{15}{61}$	$\dfrac{10}{61}$	$\dfrac{30}{61}$	$\dfrac{6}{61}$