The following is the probability distribution of X X 0 1 2 3 P( X =x) 1+ p 5 2-2 p 5 2- p 5 2 p 5 For a minimum value of p, the value of 5 E ( X ) is

Step-by-step solution

The probability distribution remains valid when all probabilities are non-negative. This requires P(X=0) = 1+p 5 0 implying p -1 , P(X=1) = 2-2p 5 0 implying p 1 , P(X=2) = 2-p 5 0 implying p 2 , and P(X=3) = 2p 5 0 implying p 0 . Combining these inequalities yields the valid range 0 p 1 . The minimum value of p is therefore 0 . Substituting p = 0 gives the distribution: P(X=0) = 1 5 , P(X=1) = 2 5 , P(X=2) = 2 5 , P(X=3) = 0 . The expected value is calculated as: E(X) = 0 1 5 + 1 2 5 + 2 2 5 + 3 0 = 2 5 + 4 5 = 6