A random variable X follows a binomial distribution B(n, p) with n=7 . If the probabilities P(X=1) , P(X=2) , and P(X=3) are in an arithmetic progression and the probability of success is not equal to the probability of failure, then the value of 3 E(X^2) is equal to

Step-by-step solution

Given that X B(n, p) with n=7 . The probabilities P(X=1) , P(X=2) , and P(X=3) are in an arithmetic progression. 2 P(X=2) = P(X=1) + P(X=3) 2 ⁷C₂ p^2 q^5 = ⁷C₁ p q^6 + ⁷C₃ p^3 q^4 2 21 p^2 q^5 = 7 p q^6 + 35 p^3 q^4 42 p^2 q^5 = 7 p q^6 + 35 p^3 q^4 Since p, q > 0 , dividing the entire equation by 7 p q^4 , we get: 6 p q = q^2 + 5 p^2 Dividing by p^2 and letting r = q p , we obtain: r^2 - 6r + 5 = 0 (r-1)(r-5) = 0 r = 1 or r = 5 If r = 1 , then q = p , which contradicts the given condition that the probability of s