Exactly three in a chosen box

Question

Ten balls are dropped independently into 5 boxes, each ball equally likely to land in any box. What is the probability that a particular box holds exactly 3 of the balls?

Accepted Answer

Each ball independently lands in the chosen box with probability \frac{1}{5} or misses with probability \frac{4}{5}. The number in that box is binomial with 10 trials and success probability \frac{1}{5}. For exactly 3 balls in the box: P = \binom{10}{3}\left(\frac15\right)^3\left(\frac45\right)^7 = 120 \cdot \frac{1}{125}\cdot\frac{16384}{78125} = \frac{1{,}966{,}080}{9{,}765{,}625} \approx 0.2013. Takeaway. A single bin's load is binomial \binom{m}{j}\left(\frac1n\right)^j\left(\frac{n-1}{n}\right)^{m-j}; here 3 balls in a chosen box happens about 20% of the time.

Exactly three in a chosen box

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