2. A grocery chain runs a prize game by giving each customer a ticket that may win a prize when a box is scratched. Printed on the ticket are the following probabilities for a customer who shops once a week.

What is the expected value of a customer's winnings in this game?

a) [tex]$\$[/tex] 0.00[tex]$
b) $[/tex]\[tex]$ 9.00$[/tex]
c) [tex]$\$[/tex] 15.00[tex]$
d) $[/tex]\[tex]$ 151.67$[/tex]

\begin{tabular}{|c|c|}
\hline
Amount Won & Probability \\
\hline
[tex]$\$[/tex] 400[tex]$ & 0.01 \\
\hline
$[/tex]\[tex]$ 40$[/tex] & 0.05 \\
\hline
[tex]$\$[/tex] 15$ & 0.20 \\
\hline
\end{tabular}

A.
B.
C.
D.



Answer :

To determine the expected value of a customer's winnings in this prize game, we'll utilize the concept of expected value from probability theory. The expected value is the sum of all possible outcomes, each multiplied by their respective probabilities. Here is a step-by-step breakdown of the calculation:

1. Identify the outcomes and their probabilities:
- Winning \[tex]$400 has a probability of 0.01. - Winning \$[/tex]40 has a probability of 0.05.
- Winning \[tex]$15 has a probability of 0.20. - The remaining probability needs to be accounted for. Since the total probability must sum to 1, the probability of not winning any prize (i.e., winning \$[/tex]0) can be calculated as follows:
[tex]\[ P(\$0) = 1 - (0.01 + 0.05 + 0.20) = 1 - 0.26 = 0.74 \][/tex]

2. Multiply each outcome by its probability:
- For \[tex]$400: \[ 400 \times 0.01 = 4.00 \] - For \$[/tex]40:
[tex]\[ 40 \times 0.05 = 2.00 \][/tex]
- For \[tex]$15: \[ 15 \times 0.20 = 3.00 \] - For \$[/tex]0:
[tex]\[ 0 \times 0.74 = 0.00 \][/tex]

3. Sum these products to find the expected value:
[tex]\[ 4.00 + 2.00 + 3.00 + 0.00 = 9.00 \][/tex]

Therefore, the expected value of a customer's winnings in this game is \[tex]$9.00. Thus, the correct answer is: \[ \boxed{\$[/tex]9.00}
\]