Consider the probability distribution ( array |r|c|c|c|c|c| X =x & 1 & 2 & 3 & 4 & 5 P ( X =x) & K & 2 K & K ^2 & 2 K & 5 K ^2 array ) Then the value of P ( X > 2) is

Step-by-step solution

The probability distribution requires P ( X =x) = 1 . With P (1) = K , P (2) = 2 K , P (3) = K ^2 , P (4) = 2 K , and P (5) = 5 K ^2 , the sum is 5 K + 6 K ^2 = 1 . Solving 6 K ^2 + 5 K - 1 = 0 yields K = -5 7 12 . Only the positive solution K = 1 6 is valid since probabilities must be non-negative. The probability P ( X > 2) is the sum P (3) + P (4) + P (5) = K ^2 + 2 K + 5 K ^2 = 6 K ^2 + 2 K . Substituting K = 1 6 gives 6 ( 1 6 )^2 + 2 ( 1 6 ) = 1 6 + 1 3 = 1 2 . The value 1 2 corresponds to option C .