Question

A game uses a single 6-sided die. To play the game, the die is rolled one time, with the following results: Even number = lose $51 or 3 = win $35 = win $11What is the expected value of the game?State your answer in terms of dollars rounded to the nearest cent (hundredth).

Answer 1

Data given:

• Possible outcomes: 1, 2, 3, 4, 5, 6

,

• Money: +$3, -$5, +$3, -$5, +$11, -$5

Each number (outcome) has a probability of:

$P=\frac{1}{6}$

Thus, the expected value ( E ) is:

$E=\frac{1}{6}\cdot(3-5+ 3- 5+ 11- 5)$$E=\frac{1}{6}\cdot2$$E=\frac{2}{6}=\frac{1}{3}\approx0.33$

Answer: 0.33