Question

in a certain game, the player rolls a standard die one time. if they roll a 1 or 2, they receive $50.00, if they roll a 3, they pay $10.00, and if they roll a 4, 5, or 6, they pay $20.00.
what is the expected value for the player in this game? round your answer to the nearest cent.

Explanation:

Step1: Define probabilities & outcomes

Probabilities: $P(1,2)=\frac{2}{6}=\frac{1}{3}$, $P(3)=\frac{1}{6}$, $P(4,5,6)=\frac{3}{6}=\frac{1}{2}$
Outcomes: $X_1=50$, $X_2=-10$, $X_3=-20$

Step2: Calculate expected value

$E(X) = P(1,2)X_1 + P(3)X_2 + P(4,5,6)X_3$
$E(X) = \frac{1}{3}(50) + \frac{1}{6}(-10) + \frac{1}{2}(-20)$

Step3: Compute each term

$\frac{50}{3} \approx 16.6667$, $\frac{-10}{6} \approx -1.6667$, $\frac{-20}{2} = -10$

Step4: Sum terms for final value

$16.6667 - 1.6667 - 10 = 5$

Answer:

$\$5.00$