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2021 AMC Fall 10A Problem 13

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2021 AMC Fall 10A Problem 13AMC 10 Step by Step
Name:                             Date: September 13, 2026
Basic Probability · difficulty 3 of 5 · about 4 min · position P11-15

Each of 66 balls is randomly and independently painted either black or white with equal probability. What is the probability that every ball is different in color from more than half of the other 55 balls?

  1. A)

    164\frac{1}{64}

  2. B)

    16\frac{1}{6}

  3. C)

    14\frac{1}{4}

  4. D)

    516\frac{5}{16}

  5. E)

    12\frac{1}{2}

Answer

D

Key insight

A black ball needs at least 3 white balls and a white ball needs at least 3 black, so the split must be exactly 3-3: C(6,3)/2^6 = 20/64.

Solution

For any one ball, the "other" balls number 55, so "more than half" means at least 33 of them have the opposite color.

Suppose kk balls are black and 6−k6 - k are white. A black ball (if any) sees 6−k6 - k white balls among the others, so we need 6−k≥36 - k \ge 3, i.e. k≤3k \le 3. A white ball (if any) sees kk black balls, so we need k≥3k \ge 3. Both colors must be present (if all six were black, each ball would differ from 00 others), so both conditions apply and k=3k = 3 exactly.

Conversely, with 33 black and 33 white, every ball differs from exactly 33 of the other 55, which is more than half. So the event is exactly "three of each color."

There are 26=642^6 = 64 equally likely colorings, of which (63)=20\binom{6}{3} = 20 have three black balls:
P=2064=516.P = \frac{20}{64} = \frac{5}{16}.

The answer is (D) 516\boxed{\textbf{(D)}\ \frac{5}{16}}.

Why this works

A condition that must hold for every ball usually collapses to a statement about the overall color count. Write the condition once for a black ball and once for a white ball; the two inequalities pin the count from both sides. Then the probability is a single binomial coefficient over 262^6.

The trap

Testing only one ball's condition and accepting 4-2 splits, or forgetting that the 'other' balls number 5, so 'more than half' means at least 3.

Common mistakes

  • Testing only one ball's condition and accepting 4-2 splits, or forgetting that the 'other' balls number 5, so 'more than half' means at least 3.
  • Counting the 33-33 colorings as (63)/2=10\binom{6}{3}/2 = 10 by treating the two colors as interchangeable, which gives 532\tfrac{5}{32}.

Techniques

Bound the quantity above/below or estimate to pin it down · Set up the equation/formula and compute; no special trick needed