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.
We do not do your homework for you. Although it might take more effort to do the work on your own, you will profit more from your effort. We will be happy to evaluate your work though.
However, I will start you out.
Out of 36 possibilities, 7 can be obtained with:
6,1
1,6
3,4
4,3
5,2
2,5
Thus his chances of winning on any one toss = 6/36 = 1/6
To find Joe's expected profit or loss, we need to calculate the probability of each outcome and then multiply it by the corresponding profit or loss.
First, let's determine the probability of getting a sum of seven when rolling two fair dice. The possible outcomes that result in a sum of seven are: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), and (6, 1). There are a total of 36 possible outcomes when rolling two dice (each die has 6 possible outcomes). So, the probability of getting a sum of seven is 6/36 or 1/6.
If Joe wins, he gets $5. So the profit in that case is $5. If Joe loses, he loses his $1. So the loss in that case is $1.
Now, let's calculate the expected profit or loss:
Expected profit or loss = (Probability of winning * Profit if won) + (Probability of losing * Loss if lost)
Probability of winning = 1/6
Profit if won = $5
Probability of losing = (Total number of outcomes - Probability of winning) / Total number of outcomes
Total number of outcomes = 36
Probability of winning = 1/6
Loss if lost = $1
Expected profit or loss = (1/6 * $5) + ((36 - 1/6) / 36 * $1)
= $5/6 + (35/36 * $1)
= $5/6 + $35/36
= $35/6 + $35/36
= ($35 * 6 + $35) / 36
= $245/36
≈ $6.81
Therefore, Joe's expected profit or loss is approximately -$6.81.