Question

Michael has 3 quarters, 2 dimes, and 3 nickels in his pocket. He randomly draws two coins from his pocket, one at a time, and they are both dimes. He says the probability of that occurring is 1
4
because 2 of the 8 coins are dimes. Is he correct? Explain.

Answer 1

Michael's reasoning is not correct. To explain why, let's go through the probability calculation step-by-step.

Firstly, we need to identify how many coins Michael has in total and how many of those are dimes:

- Quarters: 3
- Dimes: 2
- Nickels: 3

Total number of coins = 3 (quarters) + 2 (dimes) + 3 (nickels) = 8 coins.

Now, we need to calculate the probability of drawing two dimes one after another without replacement.

**Step 1: Probability of drawing the first dime**
- There are 2 dimes and 8 total coins.
- The probability of drawing a dime first is \( \frac{2}{8} = \frac{1}{4} \).

**Step 2: Probability of drawing the second dime**
- After drawing the first dime, there will be 1 dime left and now there are only 7 coins remaining (because we did not replace the first coin).
- The probability of drawing a second dime is \( \frac{1}{7} \).

**Step 3: Combined probability for both events**
- The total probability of both events happening (drawing a dime first and a dime second) is found by multiplying the probabilities of each event:

\[
P(\text{first dime}) \times P(\text{second dime | first dime}) = \frac{2}{8} \times \frac{1}{7} = \frac{1}{4} \times \frac{1}{7} = \frac{1}{28}.
\]

So the probability of drawing two dimes successively, without replacement, is \( \frac{1}{28} \).

Convincingly, Michael's statement that the probability is \( \frac{1}{4} \) is incorrect because he did not account for the changing number of coins and dimes after the first draw. The correct probability is \( \frac{1}{28} \).