7. 19 Salmon moving along the mountain stream must pass through two waterfalls. In this experiment, the probability of a salmon passing the first waterfall is $p>0$, and the probability of passing the second waterfall is $q>0$. Assume that the trials of passing the waterfalls are independent. Find the probability that, in $n$ trials, a salmon fails to pass both waterfalls, given that it fails to pass the first waterfall in $n$ trials.
[Solution] Let $A_{n}$ be the event that the salmon fails to leap the first waterfall in $n$ trials, and $B_{n}$ be the event that the salmon fails to leap both waterfalls in $n$ trials. Since the probability of failing to leap the first waterfall in one trial is $1-p$, and the trials are independent, we have
$$p\left(A_{n}\right)=(1-p)^{n}$$

The event $B_{n}$ consists of the following events: the salmon fails to leap the first waterfall in $n$ trials, or the salmon leaps the first waterfall in the $k$-th trial $(1 \leqslant k \leqslant n)$ but fails to leap the second waterfall in the subsequent $n-k$ trials. Therefore,
$$p\left(B_{n}\right)=(1-p)^{n}+\sum_{k=1}^{n}(1-p)^{k-1} p(1-q)^{n-k}$$

If $p=q$, then the right-hand side of (2) becomes
$$\begin{aligned}
p\left(B_{n}\right) & =(1-p)^{n}+\sum_{k=1}^{n}(1-p)^{n-1} p \\
& =(1-p)^{n}+n p(1-p)^{n-1}
\end{aligned}$$

If $q=1$, then the salmon leaps the second waterfall in one trial, so
$$p\left(B_{n}\right)=(1-p)^{n}+(1-p)^{n-1} p=(1-p)^{n-1}$$

Therefore, if the salmon fails to leap the first waterfall in $n$ trials, or leaps the first waterfall only in the $n$-th trial, then event $B_{n}$ occurs.

But if we define $0^{0}=1$, (4) can be derived from (2).
If $p \neq q$ and $q<1-q$, then equation (7) does not hold. If $p>q$, similarly, equation (7) does not hold.

If $p=q$, and $p\left(B_{n}\right)=0$, then from (3) we know $p=1$. Therefore, the conditional probability $p\left(A_{n} \mid B_{n}\right)$ does not exist if and only if $p=q=1$. Now let's calculate the other cases.

When $p=q<1$, from (1), (3), and (6) we get
$$p\left(A_{n} \mid B_{n}\right)=\frac{1-p}{1-p+n p}=\frac{1-p}{1+(n-1) p}$$

When $p<q=1$, from (1), (4), and (6) we get
$$p\left(A_{n} \mid B_{n}\right)=1-p$$

When $p \neq q<1$, applying (1), (5), and (6) we get
$$p\left(A_{n} \mid B_{n}\right)=\frac{(1-p)^{n}(p-q)}{p(1-q)^{n}-q(1-p)^{n}}=\frac{(p-q)\left(\frac{1-p}{1-q}\right)^{n}}{p-q\left(\frac{1-p}{1-q}\right)^{n}}$$