Let a random variable X take values 0, 1, 2, 3 with P(X = 0) = p, P(X = 1) = p, P(X = 2) = P(X = 3) and E(X²) = 2E(X). Then the value of 8p−1 is:
❓ Question Let a random variable X X take values 0 , 1 , 2 , 3 0,1,2,3 with P ( X = 0 ) = p , P ( X = 1 ) = p , P ( X = 2 ) = P ( X = 3 ) and it is given that E ( X 2 ) = 2 E ( X ) . \;E(X^2)=2E(X). Find the value of 8 p − 1. 🖼️ Question Image ✍️ Short Solution Let q = P ( X = 2 ) = P ( X = 3 ) q = P(X=2)=P(X=3) . Since total probability is 1: 2 p + 2 q = 1 ⇒ p + q = 1 2 ⇒ q = 1 2 − p . Compute the expectations: E ( X ) = 0 ⋅ p + 1 ⋅ p + 2 ⋅ q + 3 ⋅ q = p + 5 q . E(X) = 0\cdot p + 1\cdot p + 2\cdot q + 3\cdot q = p + 5q. E ( X 2 ) = 0 2 ⋅ p + 1 2 ⋅ p + 2 2 ⋅ q + 3 2 ⋅ q = p + 13 q . E(X^2) = 0^2\cdot p + 1^2\cdot p + 2^2\cdot q + 3^2\cdot q = p + 13q. Use the given relation E ( X 2 ) = 2 E ( X ) E(X^2)=2E(X) : p + 13 q = 2 ( p + 5 q ) . Simplify: p + 13 q = 2 p + 10 q ⇒ − p + 3 q = 0 ⇒ 3 q = p . Substitute q = 1 2 − p q = \tfrac12 - p into 3 q = p 3q = p : 3 ( 1 2 − p ) = p ⇒ 3 2 − 3 p = p 3\left(\tfrac12 - p\right) = p \quad\Rightarr...