AMC 10 · 2011 · #21

Grade 7 probability

Archetype Casework by Position / Vertex Type └ Sub-archetype Count of Type

conditional-probabilitycombinations-basic caseworkcomplementary-counting ↑ Prerequisites: combinations-basic

📏 Long solution 💡 3 insights

Problem

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.

Pick an answer.

(A)

$\frac{7}{11}$

(B)

$\frac{9}{13}$

(C)

$\frac{11}{15}$

(D)

$\frac{15}{19}$

(E)

$\frac{15}{16}$

AMC 10 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

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.

1STEP 1

Name the two weights

Call a genuine coin g and a counterfeit c, with g ≠ c. A pair then weighs 2g, g+c, or 2c — three different numbers.

pair weight ∈ { 2g, g+c, 2c }, g ≠ c → all three differ

2STEP 2

Decide which pairs can balance

Equal totals force the same make-up: both pairs all-genuine (2g), or both one-each (g+c). Both 2c would need 4 fakes.

balance ⇔ both pairs 2g or both pairs g+c

3STEP 3

Count the all-genuine balancing draws

All four genuine: C(8, 2)=28 ways for the first pair times C(6, 2)=15 for the second gives 420 draws.

C(8, 2)C(6, 2) = 28 × 15 = 420

4STEP 4

Count the one-counterfeit-each balancing draws

One fake per pair: 2 × 8 = 16 ways for the first pair, then 1 × 7 = 7 for the second — 112 draws.

(2 × 8)×(1 × 7) = 16 × 7 = 112

5STEP 5

Form the conditional probability

Balancing draws total 420 + 112 = 532, and 420 of them are all-genuine, so the probability is 420/532.

P = 420/(420+112) = 420/532

6STEP 6

Reduce the fraction and conclude

Both share 28: 420 = 28 × 15 and 532 = 28 × 19, so the probability is 15/19 — choice (D).

420/532 = (28 × 15)/(28 × 19) = 15/19

Answer

15/19

The answer 15/19 ≈ 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.

💡Key takeaway

When a problem says 'given that,' throw away every outcome that breaks the rule and only compare the ones that survive.

Name the two weights
Decide which pairs can balance
Count the all-genuine balancing draws
Count the one-counterfeit-each balancing draws
Form the conditional probability
Reduce the fraction and conclude