AMC 10 · 2018 · #6

Grade 7 probability

Archetype Branching State Enumeration └ Sub-archetype Constrained Sequence or Path Count

probability-basicpermutations-basic systematic-enumerationcasework ↑ Prerequisites: probability-basic

📏 Medium solution 💡 2 insights

Problem

Chips numbered 1,2,3,4,5 are drawn one at a time without replacement, stopping the moment the running sum is more than 4. Find the probability that you must draw exactly 3 chips before the sum first exceeds 4.

Pick an answer.

(A)

$frac{1}{15}$

(B)

$frac{1}{10}$

(C)

$frac{1}{6}$

(D)

$frac{1}{5}$

(E)

$frac{1}{4}$

AMC 10 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

Tool #16 (Change Focus / Count the Complement): instead of tracking the third draw, notice that needing a third draw is exactly the condition that the first two chips together sum to at most 4 — so only the first two draws matter. Tool #2 (Make a Systematic List): the first two chips form a small, listable set of pairs, so list every pair whose sum is at most 4 and count the ordered ways they can appear. Tool #3 (Eliminate Possibilities): the probability must be one of the five fractions, which lets us confirm 1/5 and rule out the others.

1STEP 1

Reframe: when is a third draw needed

A third draw is needed only when the first two chips already sum to at most 4 — otherwise you would have stopped sooner.

first chip + second chip ≤ 4

2STEP 2

List the small-sum pairs

Only two pairs of different chips sum to 4 or less: {1,2} and {1,3} — every pair without a 1 already passes 4.

1+2=3 ≤ 4, 1+3=4 ≤ 4; all other pairs ＞ 4

3STEP 3

Count favorable ordered draws

Order matters, so each pair splits into two draws — (1,2),(2,1),(1,3),(3,1) — giving 4 favorable sequences.

2 pairs × 2 orders = 4 favorable sequences

4STEP 4

Count all possible first two draws

The first two draws come out in 5 × 4 = 20 equally likely ordered ways: 5 chips, then 4 remaining.

5 × 4 = 20 total ordered sequences

5STEP 5

Form the probability

Divide favorable by total: 4/20 = 1/5, which is choice (D).

4/20 = 1/5 → (D)

Answer

frac{1}{5}

The answer 1/5 is a small but not tiny probability, which fits: needing a third draw forces a special low pair (a 1 paired with a 2 or a 3), and that is uncommon but not rare. The count is exact — 4 favorable orders out of 20 — and 4/20 reduces cleanly to 1/5, matching (D). Choices (A) 1/15 and (B) 1/10 are too small (they would need fewer than 4 favorable orders), (E) 1/4 would need 5 favorable orders, and (C) 1/6 does not divide 20 evenly — only (D) fits.

💡Key takeaway

You only need a third draw when the first two chips sum to at most 4 — that is just {1,2} or {1,3}, giving 4 ordered ways out of 20, so the probability is 1/5.

Reframe: when is a third draw needed
List the small-sum pairs
Count favorable ordered draws
Count all possible first two draws
Form the probability

Reframe: when is a third draw needed

List the small-sum pairs

Count favorable ordered draws

Count all possible first two draws

Form the probability

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