Apair of fair dice is rolled once. suppose that you lose $8 if the dice sum to 10 and win $11 if the dice sum to 11 or 12. how much should you win or lose if any other number turns up in order for the game to be fair?

Loss of $ 0.3.

Step-by-step explanation:

Since, when two dices are rolled,

Then the all possible outcomes = 36,

Also, the possible way of getting the sum of 10 are,

(4, 5), (5, 4), (5, 5),

So, the possibility of getting the sum of 10 =

Now, the possible way of getting the sum of 11 or 12 are,

(5, 6), (6, 5), (6, 6),

So, the possibility of getting the sum of 11 or 12 =

Now, the possible number of ways of getting other number= 36 - 3 - 3 = 30,

Thus, the possibility of getting other number =

Given, for getting the sum of 10 profit is -$ 8, for the sum of 11 or 12 the profit is $11,  ( '-' sign shows the loss )

Let x be the profit of getting other number,

So, the expected value of the game =

If the game is fair,

Expected value of game = 0

Hence, there should be a loss of $ 0.3 if any other number turns up in order for the game to be fair.

Alright then!

First off, is your picture REALLY necessary? Whatever haha

A mixed number is just a simplified fraction really.

16/3 = 5 1/3

Hope it helps!
Can you give me brainliest as well please? Thanks ~Samuel