The range of a random variable X is 0 , 1 ,   2 . If P X = 0 = 3 c 3 , P X = 1 = 4 c - 10 c 2 ; P X = 2 = 5 c - 1 then P 0 < X ≤ 2 =

Step-by-step solution

Random variable x   is 0 , 1 , 2   P X = 0 = 3 c 3 , P X = 1 = 4 c - 10 c 2 P X = 2 = 5 c - 1 Sum of all probabilities for a random variable is always 1 ⇒   P X = 0 + P X = 1 + P X = 2 = 1 ⇒   3 c 3 - 10 c 2 + 9 c - 1 = 1   ⇒   3 c 3 - 10 c 2 + 9 c - 2 = 0 ⇒ c - 1 c - 2 3 c - 1 = 0 Only possible value of c is 1 3   to keep probabilities between 0 and 1 . Now, P X = 0 = 3 c 3 = 3 × 1 27 = 1 9 P X = 1 = 4 c - 10 c 2 = 4 × 1 3 - 10 × 1 9 = 2 9 P