A discrete random variable X takes integer values from 1 to 5 . Its cumulative distribution function F(x) = P(X x) is given by F(x) = x^2 + x 30 for x 1, 2, 3, 4, 5 . The conditional probability P(X is a prime number X 3) is equal to

Step-by-step solution

First, we find the probability mass function P(X = x) using the relation P(X = x) = F(x) - F(x-1) , with F(0) = 0 . P(X = 1) = F(1) - F(0) = 1^2 + 1 30 - 0 = 2 30 P(X = 2) = F(2) - F(1) = 2^2 + 2 30 - 2 30 = 4 30 P(X = 3) = F(3) - F(2) = 3^2 + 3 30 - 6 30 = 6 30 P(X = 4) = F(4) - F(3) = 4^2 + 4 30 - 12 30 = 8 30 P(X = 5) = F(5) - F(4) = 5^2 + 5 30 - 20 30 = 10 30 Let A be the event that X is a prime number. The prime numbers in the given range are 2, 3, 5 . Let B be the event that X 3 , which means X 3, 4, 5 . We n