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Revision as of 09:45, 8 May 2011
Problem 21
Two counterfeit coins of equal weight are mixed with identical genuine coins. The weight of each of the counterfeit coins is different from the weight of each of the genuine coins. A pair of coins is selected at random without replacement from the coins. A second pair is selected at random without replacement from the remaining coins. The combined weight of the first pair is equal to the combined weight of the second pair. What is the probability that all selected coins are genuine?
Solution
Note that we are trying to find the conditional probability where is the coins being genuine and is the sum of the weight of the coins being equal. The only possibilities for are (g is abbreviated for genuine and c is abbreviated for counterfeit, and the pairs are the first and last two in the quadruple) . We see that happens with probability , and happens with probability , hence .
See Also
2011 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |