Key point to solve the problem:
If a probability distribution is given then as per its definition, Sum of probabilities associated with each value of random variable of given distribution is equal to 1
i.e. ∑(pi) = 1
Let,
2P(X = x1) = 3P (X = x2) = P (X = x3) = 5P(X = x4) = k (say)
∴ P(X = x1) = k/2
P(X = x2) = k/3
P(X = x3) = k
P(X = x4) = k/5
∵ ∑(pi) = 1 {∵ it is given that it is a probability distribution}

15k + 30k + 10k + 6k = 30 [ by taace LCM ]
61k = 30
k = 30/61
∴ the required probability distribution is :
