Answer :

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. ∑(p_{i}) = 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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