AMC 10 · 2011 · #21
Grade 7 probabilityTwo 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?
Pick an answer.
AMC 10 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: There are $10$ coins: $8$ genuine coins that all weigh the same, and $2$ counterfeit coins that weigh the same as each other but a different amount from a genuine coin. You draw a first pair of $2$ coins, then a second pair of $2$ coins from the $8$ that remain. You are told the two pairs weigh exactly the same. Given that fact, find the probability that all $4$ drawn coins are genuine.
Givens: $8$ genuine coins share one common weight; $2$ counterfeit coins share one common weight, different from the genuine weight; A first pair is drawn from all $10$ coins, then a second pair from the remaining $8$; The first pair and the second pair have equal combined weight; Answer choices: (A) $\frac{7}{11}$, (B) $\frac{9}{13}$, (C) $\frac{11}{15}$, (D) $\frac{15}{19}$, (E) $\frac{15}{16}$
Unknowns: The probability that all $4$ chosen coins are genuine, given that the two pairs balance
Understand
Restated: There are $10$ coins: $8$ genuine coins that all weigh the same, and $2$ counterfeit coins that weigh the same as each other but a different amount from a genuine coin. You draw a first pair of $2$ coins, then a second pair of $2$ coins from the $8$ that remain. You are told the two pairs weigh exactly the same. Given that fact, find the probability that all $4$ drawn coins are genuine.
Givens: $8$ genuine coins share one common weight; $2$ counterfeit coins share one common weight, different from the genuine weight; A first pair is drawn from all $10$ coins, then a second pair from the remaining $8$; The first pair and the second pair have equal combined weight; Answer choices: (A) $\frac{7}{11}$, (B) $\frac{9}{13}$, (C) $\frac{11}{15}$, (D) $\frac{15}{19}$, (E) $\frac{15}{16}$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #4 Introduce a Variable, #16 Change Focus / Count the Complement
The heart of the problem is deciding which pairs can possibly balance and then counting them, so Tool #2 (Make a Systematic List) is the anchor: sort every allowed outcome into a small number of clean cases and count each. Tool #4 (Introduce a Variable) sets up the counting by naming the genuine weight $g$ and counterfeit weight $c$, which reveals that a pair can only weigh $2g$, $g+c$, or $2c$. Tool #16 (Change Focus) handles the word 'given' correctly: the equal-weight statement is a condition, so we do not divide by all draws — we count only the balancing draws and ask what fraction of those are all-genuine.
Execute — Answer: D
7.SP.C.7 Step 1 Name the two weights
- Let each genuine coin weigh $g$ and each counterfeit coin weigh $c$, with $g \ne c$.
- A pair of two coins can only be one of three kinds: two genuine (weight $2g$), one of each (weight $g+c$), or two counterfeit (weight $2c$).
- Because $g \ne c$, these three totals $2g$, $g+c$, $2c$ are three different numbers.
💡 Giving the weights names turns 'equal weight' into an equation you can compare exactly.
7.SP.C.7 Step 2 Decide which pairs can balance
- The two pairs balance only if they have the same total weight.
- Since $2g$, $g+c$, and $2c$ are all different, two pairs match only when they are the same kind.
- So either both pairs are all-genuine (each weighs $2g$), or both pairs are one-genuine-one-counterfeit (each weighs $g+c$).
- Both pairs being two-counterfeit is impossible, because that would need $4$ counterfeit coins and only $2$ exist.
- This is the change of focus: from now on we only count these balancing draws.
💡 Three different pair-weights can only tie against a copy of themselves, so matching forces identical make-up.
7.SP.C.8 Step 3 Count the all-genuine balancing draws
- Here every one of the $4$ coins is genuine.
- Choose the first pair from the $8$ genuine coins: $\binom{8}{2}=\frac{8\cdot7}{2}=28$ ways.
- Then choose the second pair from the $6$ genuine coins still left: $\binom{6}{2}=\frac{6\cdot5}{2}=15$ ways.
- That gives $28 \times 15 = 420$ balancing draws that are all genuine.
💡 Two all-genuine pairs always weigh $2g=2g$, so every such draw automatically balances.
7.SP.C.8 Step 4 Count the one-counterfeit-each balancing draws
- Here each pair holds exactly one counterfeit coin, so both counterfeits are used, one per pair.
- Build the first pair: pick which counterfeit goes in it ($2$ ways) and its genuine partner ($8$ ways), giving $2 \times 8 = 16$ ways.
- The second pair must take the remaining counterfeit ($1$ way) and a genuine coin from the $7$ still left ($7$ ways).
- That is $16 \times 7 = 112$ balancing draws with a counterfeit in each pair.
💡 Splitting one counterfeit into each pair makes both pairs weigh $g+c$, so they tie.
7.SP.C.8 Step 5 Form the conditional probability
- Every balancing draw is one of these two cases, so the total number of balancing draws is $420 + 112 = 532$.
- The favorable draws — the ones with all $4$ genuine — number $420$.
- The probability we want is the favorable balancing draws over all balancing draws.
💡 Because we already threw away every non-balancing draw, we just compare counts inside the balancing world.
6.NS.B.4 Step 6 Reduce the fraction and conclude
- Simplify $\frac{420}{532}$.
- Both share the factor $28$: $420 = 28 \times 15$ and $532 = 28 \times 19$, so $\frac{420}{532} = \frac{15}{19}$.
- The probability that all four selected coins are genuine is $\frac{15}{19}$, which is answer $(D)$.
💡 Dividing top and bottom by their greatest common factor gives the fraction in lowest terms.
7.SP.C.7 Let each genuine coin weigh $g$ and each counterfeit coin weigh $c$, with $g \ne 7.SP.C.7 The two pairs balance only if they have the same total weight. Since $2g$, $g+c$ 7.SP.C.8 Here every one of the $4$ coins is genuine. Choose the first pair from the $8$ g 7.SP.C.8 Here each pair holds exactly one counterfeit coin, so both counterfeits are used 7.SP.C.8 Every balancing draw is one of these two cases, so the total number of balancing 6.NS.B.4 Simplify $\frac{420}{532}$. Both share the factor $28$: $420 = 28 \times 15$ and Review
Reasonableness: The answer $\frac{15}{19} \approx 0.79$ is a large probability, which fits: genuine coins vastly outnumber counterfeits, so most balancing draws are all-genuine. It also sits strictly between $0$ and $1$, and the counting is consistent — $420$ favorable out of $532$ total balancing draws, with $420+112=532$ accounting for every allowed case. Notice a counterfeit pair by itself ($2c$) could never balance against a genuine pair ($2g$), so those draws were correctly excluded rather than miscounted.
Alternative: Instead of counting draws, line up all $10$ coins in a row where the first two coins are pair one, the next two are pair two, and the rest are unused. Balancing fails exactly when the two counterfeits land unevenly — both in one pair, or one in a pair and one unused. Removing those cases leaves the same ratio, and the all-genuine share of the survivors is again $\frac{15}{19}$.
CCSS standards used (min grade 7)
7.SP.C.7Develop probability models and use them to find probabilities of events (Modeling a pair's weight as $2g$, $g+c$, or $2c$ and reasoning about which pairs can balance)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Counting the balancing draws case by case and forming the probability from those counts)6.NS.B.4Find greatest common factor and least common multiple of two numbers (Reducing $\frac{420}{532}$ to $\frac{15}{19}$ using the common factor $28$)
⭐ When a problem says 'given that,' throw away every outcome that breaks the rule and only compare the ones that survive.
⭐ When a problem says 'given that,' throw away every outcome that breaks the rule and only compare the ones that survive.
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