A discrete random variable X takes values in the set -1, 0, 1, 2 . The probabilities are given as P(X = -1) = a , P(X = 0) = 2a , P(X = 1) = b , and P(X = 2) = 3b , where a and b are constants. If the mean of X is 1 2 , then the conditional probability P(X 0 X 1) is equal to

Step-by-step solution

The sum of all probabilities must be 1 : P(X = -1) + P(X = 0) + P(X = 1) + P(X = 2) = 1 a + 2a + b + 3b = 1 3a + 4b = 1 ... (i) The mean (expected value) of X is given by E[X] = x_i P(x_i) : E[X] = (-1)(a) + (0)(2a) + (1)(b) + (2)(3b) = 1 2 -a + 7b = 1 2 -2a + 14b = 1 ... (ii) Solving equations (i) and (ii): Multiply (i) by 2 and (ii) by 3 : 6a + 8b = 2 -6a + 42b = 3 Adding them gives 50b = 5 b = 1 10 . Substituting b in (i): 3a + 4 10 = 1 3a = 6 10 a = 1 5 . Now, substitute a and b to find the probabilities: P(X =