A random variable X has the following probability distribution A random variable X has the following probability distribution ( array |r|c|c|c|c|c| X : & 0 & 1 & 2 & 3 & 4 P ( X ): & k & 2 k & 4 k & 2 k & k array ) then the value of ( P (1 X <4 / X 2)= ) then the value of P (1 < X < 4 / X < 2)=

Step-by-step solution

Probability Distribution: X: 0, 1, 2, 3, 4 P(X): k, 2k, 4k, 2k, k The total probability must sum to 1 : k + 2k + 4k + 2k + k = 10k = 1 , yielding k = 1 10 . The probabilities are thus P(X=0) = 1 10 , P(X=1) = 2 10 , P(X=2) = 4 10 , P(X=3) = 2 10 , P(X=4) = 1 10 . For the conditional probability P(1 X P(B) = P(X 2) = 1 10 + 2 10 + 4 10 = 7 10 . A B = 1, 2 , so P(A B) = 2 10 + 4 10 = 6 10 . By the definition of conditional probability: P(A B) = P(A B) P(B) = 6/10 7/10 = 6 7 . Final Answer: 6 7