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2019 AMC 10B Problem 17

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2019 AMC 10B Problem 17AMC 10 Step by Step
Name:                             Date: September 13, 2026
Basic Probability · difficulty 3 of 5 · about 4 min · position P16-20

A red ball and a green ball are randomly and independently tossed into bins numbered with positive integers so that for each ball, the probability that it is tossed into bin kk is 2−k2^{-k} for k=1,2,3,….k=1,2,3,\ldots. What is the probability that the red ball is tossed into a higher-numbered bin than the green ball?

  1. A)

    14\frac{1}{4}

  2. B)

    27\frac{2}{7}

  3. C)

    13\frac{1}{3}

  4. D)

    38\frac{3}{8}

  5. E)

    37\frac{3}{7}

Answer

C

Key insight

Red higher and green higher are equally likely; the only other outcome is a tie, with probability the geometric sum of 4^(-k), which is 1/3.

Solution

The two balls follow the same distribution independently, so "red higher" and "green higher" have the same probability pp. The remaining possibility is a tie:
p+p+P(tie)=1.p + p + P(\text{tie}) = 1.

Both balls land in bin kk with probability 2−k⋅2−k=4−k2^{-k} \cdot 2^{-k} = 4^{-k}, so
P(tie)=∑k=1∞4−k=1/41−1/4=13.P(\text{tie}) = \sum_{k=1}^{\infty} 4^{-k} = \frac{1/4}{1 - 1/4} = \frac13.

Therefore 2p=1−13=232p = 1 - \frac13 = \frac23 and p=13p = \frac13.

The answer is (C) 13\boxed{\textbf{(C)}\ \frac{1}{3}}.

Why this works

When two identical random quantities are compared, symmetry pairs the two strict outcomes, so only the tie probability needs computing. Ties are the easy case because they collapse a double sum into a single geometric series. Look for this "half of one minus the tie" structure whenever two independent, identically distributed objects are compared.

Alternative approach

Direct double sum: for green in bin kk, red is higher with probability ∑j>k2−j=2−k\sum_{j > k} 2^{-j} = 2^{-k}. So p=∑k≥12−k⋅2−k=∑4−k=13p = \sum_{k \ge 1} 2^{-k} \cdot 2^{-k} = \sum 4^{-k} = \frac13, the same series.

The trap

Answering 1/2 by forgetting that ties are possible, or summing 4^(-k) from k = 0 and getting 4/3.

Common mistakes

  • Answering 1/2 by forgetting that ties are possible, or summing 4^(-k) from k = 0 and getting 4/3.
  • Computing P(tie)=13P(\text{tie}) = \frac13 and then reporting it (choice C happens to coincide here, but the reasoning (1−13)/2(1 - \frac13)/2 is what generalizes).

Techniques

Count the complement and subtract from the total · Exploit symmetry to reduce work or pair up objects