In the gambling game chuck-a-luck, for a $1bet it is possible to win $1, $2, or $3 with respective probabilities 75/216,15/216,and1/216. One dollar is lost with probability 125/216. Let X equal the payoff for this game and find E(X). Note that when a bet is won, the $1 that was bet, in addition to the $1, $2, or $3 that is won, is returned to the bettor.

Answer:[tex]\frac{\textup{-17}}{\textup{216}}[/tex]  Step-by-step explanation:Given:Probability of winning $1, P($1) = [tex]\frac{\textup{75}}{\textup{216}}[/tex]Probability of winning $2, P($2) = [tex]\frac{\textup{15}}{\textup{216}}[/tex]Probability of winning $3, P($3) = [tex]\frac{\textup{1}}{\textup{216}}[/tex]Probability of losing $1, P(-$1) = [tex]\frac{\textup{125}}{\textup{216}}[/tex]Now,Expected return = ∑ ( X × (PX) )here, X is the payoff and, P(X) is the probability of winning the losing amount is depicted with the negative payoff Thus,expected return = $1 × P($1) + $2 × P($2) + $3 × P($3) + (-$1) × P(-$1)orexpected return = [tex]\frac{\textup{75}}{\textup{216}}[/tex]  +  [tex]\frac{\textup{15}}{\textup{216}}[/tex]  + [tex]\frac{\textup{1}}{\textup{216}}[/tex]  - [tex]\frac{\textup{125}}{\textup{216}}[/tex]orexpected return = [tex]\frac{(1\times75) + (2\times15) + (3\times1) - (1\times125)}{\textup{216}}[/tex]  orexpected return = [tex]\frac{\textup{-17}}{\textup{216}}[/tex]