11. Let $\xi_{1}, \ldots, \xi_{n}$ be independent Bernoulli random variables,

$$
\mathbf{P}\left\{\xi_{k}=1\right\}=p, \quad \mathbf{P}\left\{\xi_{k}=0\right\}=1-p, \quad 1 \leqslant k \leqslant n
$$

Find the conditional probability that the first one («success») appears on the $m$-th step, given that the «success» occurred only once in all $n$ steps.
Solution. Let $S_{n}:=\xi_{1}+\ldots+\xi_{n}$. The quantity $S_{n}$ has a binomial distribution with parameters $n$ and $p$. By the definition of conditional probability,

$$
\begin{aligned}
& \mathrm{P}\left\{\xi_{m}=1 \mid S_{n}=1\right\}= \\
& =\frac{\mathrm{P}\left\{\xi_{1}=\ldots=\xi_{m-1}=\xi_{m+1}=\ldots=\xi_{n}=0, \xi_{m}=1\right\}}{\mathrm{P}\left\{S_{n}=1\right\}}= \\
& \quad=\frac{p(1-p)^{n-1}}{C_{n}^{1} p(1-p)^{n-1}}=\frac{1}{n} .
\end{aligned}
$$