Exactly two of five in one urn

Question

Five balls are dropped independently into 2 urns, each ball equally likely to go into either urn. What is the probability that the left urn holds exactly 2 of the five balls?

Accepted Answer

Each ball lands in the left urn with probability \frac{1}{2}, independently, so the left urn's count is binomial with 5 trials and success probability \frac{1}{2}. For exactly 2 balls in the left urn: P = \binom{5}{2}\left(\frac12\right)^2\left(\frac12\right)^3 = 10 \cdot \frac{1}{4}\cdot\frac{1}{8} = \frac{10}{32} = \frac{5}{16} = 0.3125. With two urns the general formula collapses to \binom{5}{2}\left(\frac12\right)^5, the familiar coin-flip count. Takeaway. With two bins, a bin's load is just a fair-coin binomial \binom{m}{j}\left(\frac12\right)^m; the general occupancy formula reduces to …

Exactly two of five in one urn

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