If a discrete random variable X takes the values 1, 2, 3, 4 such that 2P(X = 1) = 3P(X = 2) = P(X = 3) = 5P(X = 4) , then P(X = 4) =

Step-by-step solution

Let P(X = 4) = k . From the given relation 2P(X = 1) = 3P(X = 2) = P(X = 3) = 5P(X = 4) , we have: P(X = 3) = 5k P(X = 2) = 5k 3 P(X = 1) = 5k 2 Since the sum of all probabilities for a discrete random variable is 1 : P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 1 5k 2 + 5k 3 + 5k + k = 1 k ( 15 6 + 10 6 + 30 6 + 6 6 ) = 1 k ( 61 6 ) = 1 k = 6 61 Therefore, P(X = 4) = 6 61 . Answer: 6 61