Answer :

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 :



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