Competition · AMC preparation · step 4 of 4

AMC 8 · 2016 · #21

Grade 7 probabilitycounting

Archetype Overlap / Complement Counting └ Sub-archetype At-Least-One Complement: 1 − P(None)

probability-basiccombinations-basiccomplementary-counting complementary-countingsystematic-enumerationeasier-related-problem ↑ Prerequisites: probability-basiccombinations-basic

📏 Medium solution 💡 4 insights

Problem

A hat holds 3 red chips and 2 green chips. We draw one chip at a time, without putting any back, and stop the instant we have drawn all 3 reds OR both greens — whichever happens first. What is the probability that the stop happens because all 3 reds came out?

Pick an answer.

(A)

$\dfrac{3}{10}$

(B)

$\dfrac{2}{5}$

(C)

$\dfrac{1}{2}$

(D)

$\dfrac{3}{5}$

(E)

$\dfrac{7}{10}$

AMC 8 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Change Focus / Count the Complement

Tracking the stopping moment directly is messy (the game can end on draw 3, 4, or 5). Tool #16 (Change Focus) gives a clean reframe: pretend we keep drawing all 5 chips. Then "all 3 reds came out first" is exactly the same as "the very last chip in the full sequence is green" — because the colour that finishes last is the colour that did NOT trigger the stop. That turns a stopping-rule problem into a one-line question: which colour sits in position 5? Tool #9 (Easier Related Problem) — listing all C(5, 2) = 10 orderings of 3R, 2G — keeps the count concrete, and Tool #2 (Systematic List) gives an order to write them in so nothing is missed.

1STEP 1

Reframe as the last chip

Ignore the stopping rule and imagine drawing all 5 chips: reds finishing first is the same event as the last chip being green.

P(reds finish first) = P(last chip is green)

2STEP 2

List every ordering

List every ordering of 3R and 2G by where the two greens sit — there are C(5, 2) = 10 equally likely orderings.

GG positions: (1,2),(1,3),(1,4),(1,5),(2,3),(2,4),(2,5),(3,4),(3,5),(4,5) → 10 orderings

3STEP 3

Count the favorable orderings

The winners are the orderings with a green in the last seat — position 5 in the pair, giving 4 winning orderings.

Favorable orderings = 4

4STEP 4

Form the probability

All 10 orderings are equally likely, so the probability is favorable over total: $\frac{4}{10}$ = $\frac{2}{5}$ , choice (B).

P(last chip is green) = 4/10 = 2/5 → (B)

Answer

2/5

Sanity-check with a one-line shortcut: each of the 5 chips is equally likely to end up in any of the 5 positions, so the chance that the chip in position 5 is green is just (greens) / (total) =

\frac{2}{5}

. That matches our answer. It is also sensible that the answer is less than

\frac{1}{2}

— the greens only need 2 draws to be "used up" while the reds need 3, so reds are slightly less likely to finish first.

\frac{2}{5}

= 0.4 fits.

💡Key takeaway

Hard-looking probability problems often shrink to one easy question — here, just "is the last chip green?" — once you change focus.

Reframe as the last chip
List every ordering
Count the favorable orderings
Form the probability

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Reframe as the last chip

List every ordering

Count the favorable orderings

Form the probability

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