2021 AMC Fall 10A Problem 13

P11-15~4 minPrint

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

Each of $6$ 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 $5$ balls?

Show solution without answering

Key insight

Each ball needs at least 3 of the other color among the other 5. That forces exactly 3 black and 3 white: 20 of 64 colorings.

Solution

What "more than half" means. Each ball has $5$ other balls. More than half of $5$ means at least $3$ . So every ball must have at least $3$ balls of the other color.
Try the possible color splits. Only the number of black balls matters. Go through the splits:

black, white	a black ball sees this many white	a white ball sees this many black	works?
$6, 0$	$0$	(no white ball)	no
$5, 1$	$1$	$5$	no
$4, 2$	$2$	$4$	no
$3, 3$	$3$	$3$	yes
$2, 4$	$4$	$2$	no
$1, 5$ and $0, 6$	same as above with colors swapped		no

Only the split $3$ black, $3$ white works. With it, every ball differs from exactly $3$ of the other $5$ .
3. Count all colorings. Each ball has $2$ choices, so there are $2^6 = 64$ equally likely colorings.
4. Count the good ones. Pick which $3$ of the $6$ balls are black: $\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$ ways.
5. Probability. $\frac{20}{64} = \frac{5}{16}$ .

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

Why this works

A rule that must hold for every ball usually turns into a rule about the totals. The black balls push the number of whites up, and the white balls push the number of blacks up. With only $6$ balls, the two pushes meet at exactly $3$ and $3$ . After that, it is one counting step over $64$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

For any one ball, the "other" balls number $5$ , so "more than half" means at least $3$ of them have the opposite color.

Suppose $k$ balls are black and $6 - k$ are white. A black ball (if any) sees $6 - k$ white balls among the others, so we need $6 - k \ge 3$ , i.e. $k \le 3$ . A white ball (if any) sees $k$ black balls, so we need $k \ge 3$ . Both colors must be present (if all six were black, each ball would differ from $0$ others), so both conditions apply and $k = 3$ exactly.

Conversely, with $3$ black and $3$ white, every ball differs from exactly $3$ of the other $5$ , which is more than half. So the event is exactly "three of each color."

There are $2^6 = 64$ equally likely colorings, of which $\binom{6}{3} = 20$ have three black balls:
$P = \frac{20}{64} = \frac{5}{16}.$

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

Review in the AoPS books

Introduction to Counting & Probability, Chapter 7: Introduction to Probability
Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Introduction to Counting & Probability, Chapter 8: Basic Probability Techniques
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques