AMC 10 · 2019 · #17

Grade 8 probability
Archetype Overlap / Complement Counting └ Sub-archetype At-Least-One Complement: 1 − P(None)
probability-basicsymmetry-argumentsequences-geometriccomplementary-counting complementary-countingpattern-recognition ↑ Prerequisites: probability-basicsequences-geometric
📏 Short solution 💡 3 insights
Problem
A red ball and a green ball are tossed independently into bins numbered 1, 2, 3, …. For each ball, the probability of landing in bin k is 2^-k. Find the probability that the red ball lands in a higher-numbered bin than the green ball.

Pick an answer.

(A)
$\frac{1}{4}$
(B)
$\frac{2}{7}$
(C)
$\frac{1}{3}$
(D)
$\frac{3}{8}$
(E)
$\frac{3}{7}$

AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Change Focus / Count the Complement

Tool #16 (Complement): direct summation Σ_g Σ_r ＞ g 2^-g · 2^-r is messy. Instead, count the tie probability P(R = G) and use symmetry. Tool #15 (Reorganize): the three events R ＞ G, R ＜ G, R = G partition the sample space — that's the right way to slice the problem. Tool #5 (Pattern): the tie probability sums a geometric series in 4^-k, an arithmetic move kids see early.

1STEP 1

Swapping red ⇔ green leaves the distribution unchanged, so P(R ＞ G) = P(R ＜ G); call this value p.

P(R ＞ G) = P(R ＜ G) = p
2STEP 2

R ＞ G, R ＜ G, R = G are disjoint and exhaust every outcome, so their probabilities sum to 1.

P(R ＞ G) + P(R ＜ G) + P(R = G) = 1
3STEP 3

Ties are easy to count: both balls land in bin k with probability 2^-k · 2^-k = 4^-k by independence.

P(R = G) = Σ_k=1^∞ 4^-k
4STEP 4

Sum the geometric series (first term 14\frac{1}{4}, ratio 14\frac{1}{4}): P(R = G) = 141−14\frac{\frac{1}{4}}{1 - \frac{1}{4}} = 13\frac{1}{3}.

P(R = G) = 141−14\frac{\frac{1}{4}}{1 - \frac{1}{4}} = 13\frac{1}{3}
5STEP 5

Substitute: 2p + 13\frac{1}{3} = 1, so p = 13\frac{1}{3} and P(R ＞ G) = 13\frac{1}{3} — choice (C).

2p + 13\frac{1}{3} = 1 → p = 13\frac{1}{3}
Answer
13\frac{1}{3}
Three outcomes (red ＞ green, red ＜ green, tie) each carry probability 13\frac{1}{3} — pleasingly symmetric. Cross-check with the first few bins: ties at (1,1), (2,2), (3,3), … have total probability 14\frac{1}{4} + 116\frac{1}{16} + 164\frac{1}{64} + … = 13\frac{1}{3}. ✓. And the red-wins events (R, G) = (2, 1), (3, 1), (3, 2), (4, 1), … summed directly Σ_g=1^∞ 2^-g Σ_r=g+1^∞ 2^-r = Σ_g=1^∞ 2^-g · 2^-g = Σ 4^-g = 13\frac{1}{3}. ✓.
💡Key takeaway

This AMC 10 problem only needs Grade 8 exponent reasoning you already know — by symmetry, red beating green and green beating red are equally likely, and ties happen 13\frac{1}{3} of the time (geometric series). So P(R ＞ G) = 1−132\frac{1 - \frac{1}{3}}{2} = 13\frac{1}{3}. The answer is (C).

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