Question
A local high school runs the following game at an annual fundraising event. In this game, marbles are randomly picked from a bag. The bag contains four red marbles, three white marbles, and one black marble. If you pick the black marble, you win $5. If you pick a white marble, you win $2. If you pick a red marble, you do not win anything. The game costs $2 to play. What are the expected winnings of the game?
Question 8 options:
$0.38
-$0.62
-$0.13
$0.50
Question 8 options:
$0.38
-$0.62
-$0.13
$0.50
Answers
GPT-4o mini
To calculate the expected winnings from the game, we first need to determine the probabilities and outcomes associated with picking each color of marble.
1. **Count the total marbles**:
- Red marbles: 4
- White marbles: 3
- Black marble: 1
Total marbles = 4 (red) + 3 (white) + 1 (black) = 8
2. **Calculate the probabilities**:
- Probability of picking a red marble \( P(R) = \frac{4}{8} = \frac{1}{2} \)
- Probability of picking a white marble \( P(W) = \frac{3}{8} \)
- Probability of picking a black marble \( P(B) = \frac{1}{8} \)
3. **Determine the outcomes**:
- Winnings for picking a red marble = $0 (but you lose the $2 to play)
- Winnings for picking a white marble = $2 (but you lose the $2 to play, net $0)
- Winnings for picking a black marble = $5 (but you lose the $2 to play, net $3)
4. **Calculate the expected values** (taking into account both the winnings and the cost of playing):
- Expected winnings from red marbles:
\[
E(R) = P(R) \times \text{Net earnings from red} = \frac{1}{2} \times (-2) = -1
\]
- Expected winnings from white marbles:
\[
E(W) = P(W) \times \text{Net earnings from white} = \frac{3}{8} \times 0 = 0
\]
- Expected winnings from black marble:
\[
E(B) = P(B) \times \text{Net earnings from black} = \frac{1}{8} \times 3 = \frac{3}{8}
\]
5. **Sum the expected values**:
\[
E(\text{game}) = E(R) + E(W) + E(B) = -1 + 0 + \frac{3}{8}
\]
To add these, convert -1 into eighths:
\[
-1 = -\frac{8}{8}
\]
So:
\[
E(\text{game}) = -\frac{8}{8} + 0 + \frac{3}{8} = -\frac{8}{8} + \frac{3}{8} = -\frac{5}{8}
\]
Thus, the expected winnings of the game is:
\[
-\frac{5}{8} \approx -0.625
\]
Since we round to two decimal places, the expected winnings are approximately -0.62.
The answer is:
**-$0.62**
1. **Count the total marbles**:
- Red marbles: 4
- White marbles: 3
- Black marble: 1
Total marbles = 4 (red) + 3 (white) + 1 (black) = 8
2. **Calculate the probabilities**:
- Probability of picking a red marble \( P(R) = \frac{4}{8} = \frac{1}{2} \)
- Probability of picking a white marble \( P(W) = \frac{3}{8} \)
- Probability of picking a black marble \( P(B) = \frac{1}{8} \)
3. **Determine the outcomes**:
- Winnings for picking a red marble = $0 (but you lose the $2 to play)
- Winnings for picking a white marble = $2 (but you lose the $2 to play, net $0)
- Winnings for picking a black marble = $5 (but you lose the $2 to play, net $3)
4. **Calculate the expected values** (taking into account both the winnings and the cost of playing):
- Expected winnings from red marbles:
\[
E(R) = P(R) \times \text{Net earnings from red} = \frac{1}{2} \times (-2) = -1
\]
- Expected winnings from white marbles:
\[
E(W) = P(W) \times \text{Net earnings from white} = \frac{3}{8} \times 0 = 0
\]
- Expected winnings from black marble:
\[
E(B) = P(B) \times \text{Net earnings from black} = \frac{1}{8} \times 3 = \frac{3}{8}
\]
5. **Sum the expected values**:
\[
E(\text{game}) = E(R) + E(W) + E(B) = -1 + 0 + \frac{3}{8}
\]
To add these, convert -1 into eighths:
\[
-1 = -\frac{8}{8}
\]
So:
\[
E(\text{game}) = -\frac{8}{8} + 0 + \frac{3}{8} = -\frac{8}{8} + \frac{3}{8} = -\frac{5}{8}
\]
Thus, the expected winnings of the game is:
\[
-\frac{5}{8} \approx -0.625
\]
Since we round to two decimal places, the expected winnings are approximately -0.62.
The answer is:
**-$0.62**