A discrete random variable X takes values 0, 1, 2, and its probability mass function satisfies the relation (x+1)P(X=x+1) = P(X=x) for all x = 0, 1, 2, , where is a non-zero constant. If P(X=1) = P(X=2) , then the probability that X takes an odd value is

Step-by-step solution

Given the recurrence relation: P(X=x+1) = x+1 P(X=x) By substituting x = 0, 1, 2, , we can express P(X=x) in terms of P(X=0) : P(X=1) = P(X=0) P(X=2) = 2 P(X=1) = ^2 2! P(X=0) In general, P(X=x) = ^x x! P(X=0) We are given that P(X=1) = P(X=2) : P(X=0) = ^2 2 P(X=0) Since 0 and P(X=0) 0 (otherwise all probabilities would be zero), we get = 2 . The sum of all probabilities must be 1 : _ x=0 ^ P(X=x) = 1 P(X=0) _ x=0 ^ 2^x x! = 1 P(X=0) e^2 = 1 P(X=0) = e⁻² We need the probability that X takes an odd value: P(X is od