Question

In a carnival​ game, a person wagers​ $2 on the roll of two dice. if the total of the two dice is​ 2, 3,​ 4, 5, or 6 then the person gets​ $4 (the​ $2 wager and​ $2 winnings). if the total of the two dice is​ 8, 9,​ 10, 11, or 12 then the person gets nothing​ (loses $2). if the total of the two dice is​ 7, the person gets​ $1.75 back​ (loses $0.25). what is the expected value of playing the game​ once?

Answer 1

Answer: a loss of 4 cents

Step-by-step explanation:

The probability of rolling a sum of 2, 3, 4, 5, or 6 is $\dfrac{15}{36}$ which earns $2.00

The probability of rolling a sum of 28, 9, 10, 11, or 12 is $\dfrac{15}{36}$ which loses $2.00

The probability of rolling a sum of 7 is $\dfrac{6}{36}$ which loses $0.25

$\bigg(\dfrac{15}{36}\times \ 2.00\bigg)+\bigg(\dfrac{15}{36}\times -\ 2.00\bigg)+\bigg(\dfrac{6}{36}\times -\ 0.25\bigg)=\boxed{-\ 0.04}$