10 次掷硬币不出现连续正面

设 $a_n$ 为长度为 $n$ 的 0-1 序列中不含连续 1 的序列数。

按最后一位分类：

• 若最后一位是 0，则前 $n-1$ 位可为任意合法序列，数量 $a_{n-1}$；
• 若最后一位是 1，则倒数第二位必须是 0，前 $n-2$ 位为任意合法序列，数量 $a_{n-2}$。

因此递推

$$a_n=a_{n-1}+a_{n-2},\quad a_1=2,\ a_2=3.$$

这就是斐波那契数列，解得 $a_{10}=144$。

总序列数为 $2^{10}=1024$，所以所求概率为

$$\frac{144}{1024}=\boxed{\frac{9}{64}}\approx 0.140625.$$

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英文解析

Let $a_n$ be the number of sequences of length $n$ consisting of 0s and 1s that do not contain consecutive 1s.

Classifying by the last digit:

• If the last digit is 0, the preceding $n-1$ digits can form any valid sequence, resulting in $a_{n-1}$ such sequences;
• If the last digit is 1, the second-to-last digit must be 0, and the preceding $n-2$ digits can form any valid sequence, resulting in $a_{n-2}$ such sequences.

Thus, the recurrence relation is:
$$a_n=a_{n-1}+a_{n-2},\quad a_1=2,\ a_2=3.$$
This is the Fibonacci sequence, yielding $a_{10}=144$ as the solution.

The total number of sequences is $2^{10}=1024$, so the required probability is
$$\frac{144}{1024}=\boxed{\frac{9}{64}}\approx 0.140625.$$