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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.
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No. Michael confused the probability for one draw with the probability for two draws.
Correct calculation (without replacement):
- P(first coin is a dime) = 2/8.
- Given the first was a dime, there is 1 dime left out of 7 coins, so P(second is a dime | first was a dime) = 1/7.
- Multiply: P(both dimes) = (2/8)·(1/7) = 2/56 = 1/28 ≈ 0.0357.
(Alternatively, choose 2 coins from 8: C(8,2)=28 possible pairs and only C(2,2)=1 favorable pair of two dimes, so 1/28.)
If he had replaced the first coin before drawing again (with replacement), the draws would be independent and the probability would be (2/8)·(2/8)=1/16.
Correct calculation (without replacement):
- P(first coin is a dime) = 2/8.
- Given the first was a dime, there is 1 dime left out of 7 coins, so P(second is a dime | first was a dime) = 1/7.
- Multiply: P(both dimes) = (2/8)·(1/7) = 2/56 = 1/28 ≈ 0.0357.
(Alternatively, choose 2 coins from 8: C(8,2)=28 possible pairs and only C(2,2)=1 favorable pair of two dimes, so 1/28.)
If he had replaced the first coin before drawing again (with replacement), the draws would be independent and the probability would be (2/8)·(2/8)=1/16.
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