Question: A bag contains 3 gold marbles, 10 silver marbles, and 26 black marbles. Someone offers to play this game: You randomly select one marble from the bag. If it is gold, you win $4. If it is silver, you win $3. If it is black, you lose $1.
What is your expected value if you play this game?
1. **State the problem:** We want to find the expected value of the game where you draw one marble from a bag containing 3 gold, 10 silver, and 26 black marbles.
2. **Total number of marbles:** $$3 + 10 + 26 = 39$$
3. **Probabilities of each outcome:**
- Probability of gold marble: $$\frac{3}{39}$$
- Probability of silver marble: $$\frac{10}{39}$$
- Probability of black marble: $$\frac{26}{39}$$
4. **Values associated with each outcome:**
- Gold marble win: $$4$$
- Silver marble win: $$3$$
- Black marble loss: $$-1$$
5. **Expected value formula:**
$$
E = P(\text{gold}) \times \text{value}_{\text{gold}} + P(\text{silver}) \times \text{value}_{\text{silver}} + P(\text{black}) \times \text{value}_{\text{black}}
$$
6. **Substitute values:**
$$
E = \frac{3}{39} \times 4 + \frac{10}{39} \times 3 + \frac{26}{39} \times (-1)
$$
7. **Calculate each term:**
$$
\frac{3}{39} \times 4 = \frac{12}{39}
$$
$$
\frac{10}{39} \times 3 = \frac{30}{39}
$$
$$
\frac{26}{39} \times (-1) = -\frac{26}{39}
$$
8. **Sum the terms:**
$$
E = \frac{12}{39} + \frac{30}{39} - \frac{26}{39} = \frac{12 + 30 - 26}{39} = \frac{16}{39}
$$
9. **Simplify fraction if possible:** 16 and 39 have no common factors other than 1, so fraction is simplified.
10. **Convert to decimal:**
$$
E \approx 0.4103
$$
**Final answer:** The expected value of playing this game is approximately $$0.41$$, meaning on average you win about 41 cents per game.