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 $0.75 back (loses $0.25). what is the expected value of playing the game once?

The expected value of playing the game once is -$0.04

Step-by-step explanation:

Consider the provided information.

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).

The total number of outcomes are:

(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)

(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)

(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)

(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)

(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)

(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)

The probability of getting sum 2 is

The probability of getting sum 3 is

The probability of getting sum 4 is

The probability of getting sum 5 is

The probability of getting sum 6 is

Similarly, the probability of getting sum 8, 9, 10, 11, or 12 is:

The probability of P(7) is

If the total of the two dice is 2, 3, 4, 5, or 6 then the person gets $4.

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 $0.75 back (loses $0.25).

Thus the required expected value is:

Hence, the expected value of playing the game once is -$0.04

5.25 ounces

step-by-step explanation:

divided 3/4 and 6