Question

question number 7. (10.00 points) suppose you want to play a carnival game that costs 4 dollars each time you play. if you win, you get $100. the probability of winning is 3/100. what is the expected value of the amount the carnival stands to gain? 1.20 -1.00 1.00 0.90 3.00 none of the above

Explanation:

Step1: Define the two - case scenarios

The carnival stand gains $4$ dollars when the player loses and loses $(100 - 4)=96$ dollars when the player wins. The probability of the player losing is $p_1 = 1-\frac{3}{100}=\frac{97}{100}$, and the probability of the player winning is $p_2=\frac{3}{100}$.

Step2: Use the expected - value formula

The expected - value formula for a discrete random variable is $E(X)=\sum_{i}x_ip_i$. Here, $x_1 = 4$ (amount gained when player loses) and $x_2=-96$ (amount lost when player wins). So $E(X)=4\times\frac{97}{100}+(-96)\times\frac{3}{100}$.

Step3: Calculate the expected value

First, calculate $4\times\frac{97}{100}=\frac{388}{100}$ and $(-96)\times\frac{3}{100}=-\frac{288}{100}$. Then $E(X)=\frac{388 - 288}{100}=\frac{100}{100}=1.00$.

Answer:

1.00