Question

Monica tosses a fair 6-sided die. If the roll is a prime number, then she wins that amount of dollars (so that, for example, if she rolls 3, then she wins 3 dollars). If the roll is composite, she wins nothing. Otherwise, she loses 3 dollars. What is the expected value of her winnings on one die toss? Express your answer as a dollar value to the nearest cent.

Answer 1

Expected win = $15.17 to the nearest cent.

Step-by-step explanation:

Expected win = (1/6)*(1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2) = 15.167 or $15.17 to the nearest cent.

Answer 2

Answer: 1.17

The positive expected value means that Monica wins 1.17 dollars on average for each roll.

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Step-by-step explanation:

The sample space is {1,2,3,4,5,6} to represent the outcomes of the die.

Of that list, {2,3,5} are prime numbers. Note that 1 is not a prime number. It's not composite either. So Monica will lose money when she rolls a 1. If she rolls {4,6} then she wins 0 dollars.

For any single roll, the probability of landing on that value is 1/6

Multiply 1/6 by each winning amount. Then add up the products

(1/6)*(-3) + (1/6)*(2) + (1/6)*(3) + (1/6)*(0) + (1/6)*(5) + (1/6)*(0)

-3/6 + 2/6 + 3/6 + 5/6

(-3+2+3+5)/6

7/6

1.17 approximately

Monica will get about 1.17 dollars per roll on average. She only loses money on average if the expected value was negative.