A random variable X takes the values 0, 1, 2 and 3 such that:

P(X = 0) = P(X>0) = P(X<0);


P(X = -3) = P(X = -2) = P(X = -1);


P (X = 1) = P(X = 2) = P(X = 3).


Obtain the probability distribution of X.

The 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 a random variable of given distribution is equal to 1


i.e. ∑(pi) = 1


Let, P(X = 0) = k


As sum of probabilities associated with each random variable is 1


∴ P(X<0) + P(X = 0) + P(X>0) = 1


k + k + k = 1 {∵ P(X = 0) = P(X>0) = P(X<0)}


3k = 1


∴ k = 1/3


Thus


P(X<0) = 1/3


P(X = -3) + P(X = -2) + P(X = -1) = 1/3


m+m+m = 1/3 {∵ P(X = -3) = P(X = -2) = P(X = -1) = m (say) }


m = 1/9


Similarly,


P(X>0) = 1/3


P (X = 1) + P(X = 2) + P(X = 3) = 1/3


n+n+n = 1/3 {∵ P(X = 3) = P(X = 2) = P(X = 1) = n (say) }


∴ n = 1/9


∴ the required probability distribution is :



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