SOLUTION: Joe is playing a game of chance at the hibiscus festival, costing $1 for each game. In the game two fair dice are rolled and the sum of the numbers that turned up is found. If the

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Question 1138185: Joe is playing a game of chance at the hibiscus festival, costing $1 for each game. In the game two fair dice are rolled and the sum of the numbers that turned up is found. If the sum is seven, then Joe wins $5. Otherwise loses his money. Joe play the game 15 times. Find his expected profit or lose.
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


The probability of rolling a 7 is 1/6; the profit is -$1 + $5 = $4.

The probability of rolling any other sum is 5/6; the proft is -$1.

The expected profit for one game is



If he plays the game 15 times, his expected profit is



ANSWER: his expected profit playing the game 15 times is -$2.50 -- i.e., a loss of $2.50.

Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.
When two fair dice are rolled, there are 36 possible outcomes, each with the probability of  .


These outcomes are the pairs of integer numbers (i,j), where i and j take independently the values from 1 to 6.


The number of such pairs with the sum i+j = 7 is  6

( the pairs are (1,6), (2,5), (3,4), (4,3), (5,2) and (6,1) ).


So, the probability to get the sum of 7 at each roll of a pair of dice is   = .


Mathematical expectation of the amount of money to win at each roll of a pair of dice is   dollars,
but Joe should pay one dollar for each roll,

so his expected profit at each roll of a pair of dice is


     =  =   of a dollar.


Thus, statistically, Joe should expect the loss   of a dollar in each roll of a pair of dice.


Making 15 rolls in the game, Joe should expect to lose   dollars =  dollars = $2.50.     ANSWER


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