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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?

Respuesta :

tramserran

Answer: a loss of 4 cents

Step-by-step explanation:

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

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

The 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]

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