Exactly three of twelve in one bin

Question

Twelve balls are thrown independently into 4 boxes, each ball equally likely to land in any box. What is the probability that a chosen box holds exactly 3 of the twelve balls?

Accepted Answer

Each ball independently lands in the chosen box with probability \frac{1}{4} or misses with probability \frac{3}{4}. The box's count is binomial with 12 trials and success probability \frac{1}{4} (note the average load is 12/4 = 3). For exactly 3 balls in the box: P = \binom{12}{3}\left(\frac14\right)^3\left(\frac34\right)^9 = 220 \cdot \frac{1}{64}\cdot\frac{19683}{262144} = \frac{4{,}330{,}260}{16{,}777{,}216} \approx 0.2581. Even at the most likely load (3, the mean), the exact count happens only about a quarter of the time. Takeaway. A bin's load is binomial; the single most probable coun…

Exactly three of twelve in one bin

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