A discrete random variable X has the probability mass function P(X=x) = c(x+1) ^x for x = 0, 1, 2, , where c and are positive constants. If 2P(X=1) = 3P(X=2) , then the value of P(X 2) is equal to

Step-by-step solution

Given 2P(X=1) = 3P(X=2) 2[c(1+1) ^1] = 3[c(2+1) ^2] 4c = 9c ^2 Since c and are positive constants, = 4 9 . The sum of all probabilities must be 1: _ x=0 ^ P(X=x) = 1 c _ x=0 ^ (x+1) ^x = 1 The series is an arithmetico-geometric progression: S = 1 + 2 + 3 ^2 + = 1 (1- )^2 c 1 (1- )^2 = 1 c = (1- )^2 = (1 - 4 9 )^2 = 25 81 We need to find P(X 2) : P(X 2) = 1 - P(X=0) - P(X=1) P(X=0) = c(1) ^0 = c = 25 81 = 225 729 P(X=1) = c(2) ^1 = 2 25 81 4 9 = 200 729 P(X 2) = 1 - ( 225 729 + 200 729 ) = 1 - 425 729 = 304 729 Answ