连续抛硬币

我们要求前 3 次抛掷中的正面数等于后 2 次抛掷中的正面数。

对 $r = 0, 1, 2$，两边都有可能相等。

概率为 $$\sum_{r=0}^2{P(\text{前 3 次有 } r \text{ 个正面})} \cdot P(\text{后 2 次有 } r \text{ 个正面})$$

$$P(\text{前 3 次有 } r \text{ 个正面}) = \frac{3\choose r}{2^3}, \quad P(\text{后 2 次有 } r \text{ 个正面}) = \frac{2\choose r}{2^2}$$

因此

$$P = \sum_{r=0}^2 \frac{{3}\choose{r} }{8} \cdot \frac{{2}\choose{r} }{4} = \frac{1 + 6 + 3}{32} = \boxed{\frac{5}{16}}$$

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Original Explanation

We want the # of heads in the first three flips to equal the # of heads in the last 2 flips.

For $r = 0, 1, 2$ the events can match.

Probability = $\sum_{r=0}^2{P(\text{first 3 have} \ r}) \cdot P(\text{last 2 have} \ r)$

$$P(\text{first 3 have} \ r) = \frac{3\choose r}{2^3}, \ P(\text{last 2 have} \ r) = \frac{2\choose r}{2^2}$$

So

$$P = \sum_{r=0}^2 \frac{{3}\choose{r} }{8} \cdot \frac{{2}\choose{r} }{4} = \frac{1 + 6 + 3}{32} = \boxed{\frac{5}{16}}$$