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 top hat contains 3 red chips and 2 green chips. Chips are drawn randomly, one at a time without replacement, until all 3 of the reds are drawn or until both green chips are drawn. What is the probability that the 3 reds are drawn?

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.

Try it yourself first — the explanation is most useful after you’ve attempted it.

View mode:

Toolkit + CCSS Solution

Understand

Restated: 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?

Givens: $3$ red chips and $2$ green chips in the hat ($5$ chips total); Chips are drawn one at a time, without replacement; Drawing stops as soon as all $3$ reds are out, OR as soon as both greens are out; Answer choices: (A) $\tfrac{3}{10}$, (B) $\tfrac{2}{5}$, (C) $\tfrac{1}{2}$, (D) $\tfrac{3}{5}$, (E) $\tfrac{7}{10}$

Unknowns: The probability that the stopping rule is triggered by the $3$rd red (not by the $2$nd green)

Understand

Plan

Primary tool: #16 Change Focus / Count the Complement

Secondary: #9 Solve an Easier Related Problem, #2 Make a Systematic List

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 $\binom{5}{2} = 10$ orderings of $3$R, $2$G — keeps the count concrete, and Tool #2 (Systematic List) gives an order to write them in so nothing is missed.

Execute — Answer: B

#16 Change Focus / Count the Complement 7.SP.C.7 Step 1

Reframe with Tool #16.
Imagine we ignore the stopping rule and keep drawing until all $5$ chips are out.
The chip we draw last is whichever colour was NOT used up first.
So "reds get used up first" $\iff$ "the last chip in the full sequence is green".
This trick turns the problem into a much simpler question about position $5$.

$$P(\text{reds finish first}) = P(\text{last chip is green})$$

💡 If the reds run out before the greens, then at least one green chip is still sitting in the hat at the stopping moment — and if we kept drawing, that leftover green would be the very last one out.

#2 Make a Systematic List 7.SP.C.8 Step 2

Use Tool #2 to list every distinct ordering of $3$R and $2$G in $5$ positions.
We order by where the two G's sit (positions of G in increasing order).
There are $\binom{5}{2} = 10$ orderings.

$$\text{GG positions: } (1,2),(1,3),(1,4),(1,5),(2,3),(2,4),(2,5),(3,4),(3,5),(4,5) \Rightarrow 10 \text{ orderings}$$

💡 Choosing which $2$ of the $5$ positions are green fixes the whole arrangement, so $\binom{5}{2}$ counts every ordering exactly once.

#9 Solve an Easier Related Problem 7.SP.C.8 Step 3

Use Tool #9: from the list, the orderings whose LAST chip is green are exactly the ones where position $5$ is in our G-pair.
Those are $(1,5),(2,5),(3,5),(4,5)$ — $4$ orderings.

$$\text{Favorable orderings} = 4$$

💡 Listing the actual $10$ cases makes the count concrete: $4$ of them put a G in the last seat.

#16 Change Focus / Count the Complement 7.SP.C.7 Step 4

All $10$ orderings are equally likely because every draw is uniform without replacement.
The probability is the favorable count over the total count.

$$P(\text{last chip is green}) = \dfrac{4}{10} = \dfrac{2}{5} \;\Rightarrow\; \textbf{(B)}$$

💡 Equally-likely outcomes mean probability is just (good cases) $\div$ (all cases) — the foundation of Grade 7 theoretical probability.

[1] #16 7.SP.C.7 Reframe with Tool #16. Imagine we ignore the stopping rule and keep drawing unti

[2] #2 7.SP.C.8 Use Tool #2 to list every distinct ordering of $3$R and $2$G in $5$ positions. W

[3] #9 7.SP.C.8 Use Tool #9: from the list, the orderings whose LAST chip is green are exactly t

[4] #16 7.SP.C.7 All $10$ orderings are equally likely because every draw is uniform without repl

Review

Reasonableness: 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) $= 2/5$. That matches our answer. It is also sensible that the answer is less than $\tfrac{1}{2}$ — the greens only need $2$ draws to be "used up" while the reds need $3$, so reds are slightly less likely to finish first. $\tfrac{2}{5} = 0.4$ fits.

Alternative: Tool #2 (Systematic List) on its own: list the $10$ orderings of $3$R, $2$G and for each, mark whether the $3$rd R appears before the $2$nd G. Orderings ending in G (there are $4$): RRRGG, RRGRG, RGRRG, GRRRG — each ends with G and is a win. Orderings ending in R (there are $6$) all see the $2$nd G appear before the $3$rd R, so each is a loss. $4$ wins out of $10$ $= \tfrac{2}{5}$, same answer.

CCSS standards used (min grade 7)

7.SP.C.7 Develop a probability model and use it to find probabilities of events (Setting up an equally-likely model on the $10$ orderings of $3$R, $2$G and computing the probability of a stopping event as (favorable orderings) $/$ (total orderings).)
7.SP.C.8 Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation (Listing all $\binom{5}{2} = 10$ orderings of $3$R and $2$G and counting how many end with G to get the $4$ favorable cases.)

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

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 7)

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