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.

Question

Mathematics

Guest

20 February, 06:21

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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Answers (2)

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Jacey Kirk · Answer 1

Yes, because there are eight coins total and 2 are dimes and the reduced version of 2/8 is 1/4.

Deven · Answer 2

In order to get the correct answer, you have to find the percentages necessary to solve the problem.

Michael claims that the probability is 25%, but I shall disprove him. >=D

There are 8 coins total. 2 of them are dimes. The first time he draws, the probability is 25%.

The next time he draws from his pocket, there are only 7 coins and only one is a dime. The chance of him pulling that one dime is 1 / 7 or 14%.

If we average the two percentages (25 + (14/2)), we get 19.5%.

19.5% is your answer, not 25%.