Question

a game is played using one die. if the die is rolled and shows 4, the player wins $50. if the die shows any number other than 4, the player wins nothing. complete parts (a) through (b) below.
a. if there is a charge of $10 to play the game, what is the games expected value?
$\square$ (round to the nearest cent.)

Explanation:

Step1: Determine probabilities

A die has 6 faces. Probability of rolling a 4: $P(4) = \frac{1}{6}$. Probability of not rolling a 4: $P(\text{not }4) = \frac{5}{6}$.

Step2: Define outcomes

If roll 4: Net gain = $50 - 10 = 40$ (since cost is $10$). If not roll 4: Net gain = $0 - 10 = -10$.

Step3: Calculate expected value

Expected value $E = P(4) \times 40 + P(\text{not }4) \times (-10)$
$E = \frac{1}{6} \times 40 + \frac{5}{6} \times (-10)$
$E = \frac{40}{6} - \frac{50}{6}$
$E = \frac{40 - 50}{6} = \frac{-10}{6} \approx -1.67$

Answer:

\$-1.67