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2011 AMC 10A Problem 21

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2011 AMC 10A Problem 21AMC 10 Step by Step
Name:                             Date: September 13, 2026
Conditional Probability & States · difficulty 4 of 5 · about 6 min · position P21-25

Two counterfeit coins of equal weight are mixed with 8 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 10 coins. A second pair is selected at random without replacement from the remaining 8 coins. The combined weight of the first pair is equal to the combined weight of the second pair. What is the probability that all 4 selected coins are genuine?

  1. A)

    711\frac{7}{11}

  2. B)

    913\frac{9}{13}

  3. C)

    1115\frac{11}{15}

  4. D)

    1519\frac{15}{19}

  5. E)

    1516\frac{15}{16}

Answer

D

Key insight

The pairs weigh the same only when both are all genuine or each contains exactly one counterfeit, so compare just those two counts.

Solution

Every way of choosing a first pair and then a second pair is equally likely: there are (102)(82)=45⋅28=1260\binom{10}{2}\binom{8}{2} = 45 \cdot 28 = 1260 of them.

Let gg and c≠gc \ne g be the genuine and counterfeit weights. A pair weighs 2g2g, g+cg + c, or 2c2c according to how many counterfeits it holds, and these three values are all different. So the two pairs balance exactly when they contain the same number of counterfeit coins. With only two counterfeits in the box, that leaves two cases.

Case 1: no counterfeits at all. Choose two genuine coins, then two more: (82)(62)=28⋅15=420\binom{8}{2}\binom{6}{2} = 28 \cdot 15 = 420 ways.

Case 2: one counterfeit in each pair. First pair: one of 22 counterfeits with one of 88 genuine coins, 1616 ways. Second pair: the remaining counterfeit with one of the 77 remaining genuine coins, 77 ways. Total 16⋅7=11216 \cdot 7 = 112.

The condition "the weights match" restricts us to 420+112=532420 + 112 = 532 equally likely outcomes, of which the 420420 in Case 1 are all-genuine. The probability is
420532=105133=1519.\frac{420}{532} = \frac{105}{133} = \frac{15}{19}.

The answer is (D) 1519\boxed{\textbf{(D)}\ \frac{15}{19}}.

Why this works

Conditioning means throwing away every outcome that violates the given information and re-normalizing over what remains. Here the given information ("weights match") is a statement about counterfeit counts, so a short case split identifies the surviving outcomes, and the answer is a ratio of two counts. Look for the hidden equivalence that turns a physical condition into a combinatorial one.

Alternative approach

Work with probabilities instead of counts. Both pairs genuine: 2845⋅1528=13\frac{28}{45}\cdot\frac{15}{28} = \frac{1}{3}. One counterfeit in each pair: 1645⋅728=445\frac{16}{45}\cdot\frac{7}{28} = \frac{4}{45}. Conditional probability: 1/31/3+4/45=1515+4=1519\frac{1/3}{1/3 + 4/45} = \frac{15}{15 + 4} = \frac{15}{19}.

The trap

Reporting the unconditional probability that all four coins are genuine, 420/1260 = 1/3, instead of restricting to the equal-weight outcomes.

Common mistakes

  • Reporting the unconditional probability that all four coins are genuine, 420/1260 = 1/3, instead of restricting to the equal-weight outcomes.
  • Forgetting the mixed case entirely (or counting it as 16⋅1616 \cdot 16 by not removing the used counterfeit), which changes the denominator.
  • Treating the pair "two counterfeits" versus "two genuine" as balanced; 2c≠2g2c \ne 2g because c≠gc \ne g.

Techniques

Split into exhaustive cases and handle each · Set up the equation/formula and compute; no special trick needed