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Suppose that you roll a pair of honest dice. If you roll a total of​ 6, you win​ $18; if you roll a total of​ 11, you win​ $72; if you roll any other​ total, you lose​ $9. Let X denote the payoff in a single play of this game. Find the expected value of a play of this game. Round your answer to the nearest penny.A.) -​ $0.50B.) −​ $1.00C.) ​$0D.) ​$0.75E.) −​ $0.75

Respuesta :

LammettHash

You can roll a total of 6 by rolling (1, 5), (2, 4), (3, 3), (4, 2), or (5, 1). There are 36 total possible rolls that can be obtained, so you roll a total of 6 with probability 5/36.

You can roll a total of 11 by rolling (5, 6) or (6, 5), hence with probability 2/36 = 1/18.

Any of the other remaining 29 outcomes occurs with probability 29/36.

The expected value of your winnings is

$18*5/36 + $72*1/18 - $9*29/36 = -$0.75

so the answer is E.

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