2011 AMC 10A Problem 21

P21-25~6 minPrint

Conditional Probability & States Basic Counting Split into exhaustive cases and handle each Set up the equation/formula and compute; no special trick needed

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?

Show solution without answering

Key insight

The two pairs weigh the same only when both are all real coins or each holds exactly one fake. Count those two kinds of outcomes and compare.

Solution

We are told the weights matched. So we only look at the ways of picking coins where that happens, and ask what fraction of those have no fakes.

1. What makes two pairs weigh the same? Call a real coin's weight $g$ and a fake's weight $f$ (different from $g$ ). A pair with no fakes weighs $2g$ , a pair with one fake weighs $g + f$ , and a pair with two fakes weighs $2f$ . These three weights are all different. So two pairs balance exactly when they contain the same number of fakes.
2. There are only $2$ fakes, so both pairs having $2$ fakes is impossible. That leaves two cases: no fakes in either pair, or one fake in each pair.
3. Count every way to pick. Picking $2$ coins from $10$ can be done in $\frac{10 \times 9}{2} = 45$ ways. Then picking $2$ from the remaining $8$ can be done in $\frac{8 \times 7}{2} = 28$ ways. All $45 \times 28 = 1260$ ordered choices are equally likely.
4. Case 1: all four coins real. First pair: $2$ of the $8$ real coins, $\frac{8 \times 7}{2} = 28$ ways. Second pair: $2$ of the $6$ real coins left, $\frac{6 \times 5}{2} = 15$ ways. Total $28 \times 15 = 420$ .
5. Case 2: one fake in each pair. First pair: one of $2$ fakes and one of $8$ real coins, $2 \times 8 = 16$ ways. Second pair: the other fake and one of the $7$ real coins left, $7$ ways. Total $16 \times 7 = 112$ .
6. Answer the question. The weights match in $420 + 112 = 532$ of the equally likely picks, and $420$ of those are all real. So the probability is
$\frac{420}{532}.$
Divide top and bottom by $4$ : $\frac{105}{133}$ . Divide by $7$ : $\frac{15}{19}$ .

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

Why this works

"Given that the weights match" means: cross out every outcome where they don't, and work only with what is left. The key step is turning the weight condition into a counting condition (same number of fakes in each pair). After that, the answer is just the number of good outcomes divided by the number of outcomes still allowed.

Alternative approach

Use probabilities instead of counts. The chance both pairs are all real is $\frac{28}{45} \times \frac{15}{28} = \frac{15}{45} = \frac{1}{3}$ . The chance each pair has one fake is $\frac{16}{45} \times \frac{7}{28} = \frac{4}{45}$ . Of the times the weights match, the fraction that are all real is
$\frac{\frac{1}{3}}{\frac{1}{3} + \frac{4}{45}} = \frac{\frac{15}{45}}{\frac{19}{45}} = \frac{15}{19}.$

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Let $g$ and $c \ne g$ be the genuine and counterfeit weights. A pair weighs $2g$ , $g + c$ , or $2c$ 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: $\binom{8}{2}\binom{6}{2} = 28 \cdot 15 = 420$ ways.

Case 2: one counterfeit in each pair. First pair: one of $2$ counterfeits with one of $8$ genuine coins, $16$ ways. Second pair: the remaining counterfeit with one of the $7$ remaining genuine coins, $7$ ways. Total $16 \cdot 7 = 112$ .

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

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

Review in the AoPS books

Introduction to Counting & Probability, Chapter 8: Basic Probability Techniques (§8.5)
Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.2)
Intermediate Counting & Probability, Chapter 11: Conditional Probability