2011 AMC 10A Problem 21

Below is the professionally curated solution for Problem 21 of the 2011 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2011 AMC 10A solutions, or check the answer key.

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Concepts:conditional probabilitycombinationscasework

Difficulty rating: 1990

21.

Two counterfeit coins of equal weight are mixed with 88 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 1010 coins. A second pair is selected at random without replacement from the remaining 88 coins. The combined weight of the first pair is equal to the combined weight of the second pair. What is the probability that all 44 selected coins are genuine?

711\dfrac{7}{11}

913\dfrac{9}{13}

1115\dfrac{11}{15}

1519\dfrac{15}{19}

1516\dfrac{15}{16}

Solution:

There are two cases: either both selected pairs contain only genuine coins or each selected pair has one counterfeit coin.

For the first case, there are (82)=28\binom{8}{2} = 28 ways to choose the coins for the first pair and (62)=15\binom{6}{2} = 15 choices for the second pair.

We also have to divide by 22 since we can swap the pairs. This gives us 2815÷2=210 28 \cdot 15 \div 2 = 210 configurations for this case.

For the second case, there are (82)=28\binom{8}{2} = 28 ways to choose the non-counterfeit coins. There is only one choice for the counterfeit coins.

There are two ways to create the two pairs, two choices for which counterfeit coin goes with a genuine coin.

This means that there are 282=5628 \cdot 2 = 56 configurations for this case.

The desired probability is then 210210+56=210266=1519. \dfrac{210}{210 + 56} = \dfrac{210}{266} = \dfrac{15}{19}.

Thus, D is the correct answer.

Problem 21 in Other Years