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?
Accepted Solution
A:
Answer: a loss of 4 centsStep-by-step explanation:The probability of rolling a sum of 2, 3, 4, 5, or 6 is [tex]\dfrac{15}{36}[/tex] which earns $2.00The probability of rolling a sum of 28, 9, 10, 11, or 12 is [tex]\dfrac{15}{36}[/tex] which loses $2.00The probability of rolling a sum of 7 is [tex]\dfrac{6}{36}[/tex] which loses $0.25[tex]\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}[/tex]