A random variable X takes values 0,1,2,3 ,... with probabilities. P ( X =x)= k (x+1) ( 1 2 )^x, k is a constant, then P ( X =1)=

Step-by-step solution

Normalization yields the constant k . For a discrete random variable, the probabilities sum to 1: _ x=0 ^ P(X=x) = 1 . The infinite series _ x=0 ^ (x+1) ( 1 2 )^x is evaluated by differentiating the geometric series sum _ x=0 ^ r^x = 1 1-r with respect to r , yielding d dr ( 1 1-r ) = 1 (1-r)^2 . Substituting r = 1 2 , the sum becomes 1 ( 1 2 )^2 = 4 . Then, k 4 = 1 implies k = 1 4 . The desired probability is P(X=1) = 1 4 (1+1) ( 1 2 )^1 = 1 4 . This result corresponds to option C .