Probability of traversing a network of bridges This is problem 32 from this wonderful site here: https://www.probabilitycourse.com/chapter1/1_5_0_chapter1_problems.php :

You would like to go from point $A$ to point $B$. There are $5$ bridges on different branches of the river.



Bridge $i$ is open with probability $P_i$. Let $A$ be the event that there is a path from $A$ to $B$ and let $B_k$ be the event that $k$-th bridge is open.


Find $P(A)$
Find $P(B_3\ |\ A)$


My approach so far is this: $P(A)=P(A\ |\ B_3)P_3 + P(A\ |\ B_3^c)(1-P_3)$. Focusing on $P(A\ |\ B_3^c)$, we see that it's possible to cross from $A$ to $B$ only via the upper or lower route, so both bridges alone either or both routes should be open. Event $A$ becomes $(B_1\cap B_4)\cup(B_2\cap B_5)$, so

$$P(A\ |\ B_3^c)=P((B_1, B_4)\cup(B_2, B_5))=P(B_1, B_4)+P(B_2, B_5)-P(B_1, B_4, B_2, B_5)
\\=P_1P_4+P_2P_5-P_1P_2P_4P_5
$$

If bridge $3$ is open, then either of bridge $1$ or $2$ should be open, and either of bridge $4$ or $5$ should be open, so

$$P(A\ |\ B_3)=P((B_1\cup B_2)\cap(B_4\cup B_5))=P(B_1\cup B_2)+P(B_4\cup B_5)-P(B_1\cup B_2\cup B_4\cup B_5)
\\=P(B_1)+P(B_2)-P(B_1,B_2)+P(B_4)+P(B_5)-P(B_4,B_5)-[P(B_1)+P(B_2)+P(B_4)+P(B_5)-P(B_1,B_2)-P(B_1,B_4)-P(B_1,B_5)-P(B_2,B_4)-P(B_2,B_5)-P(B_4,B_5)+P(B_1,B_2,B_4)+P(B_2,B_4,B_5)+P(B_1,B_4,B_5)+P(B_1,B_2,B_5)-P(B_1,B_2,B_4,B_5)]
\\=P_1P_4+P_2P_4+P_1P_5+P_2P_5+P_1P_2P_4P_5\bigg(1-\frac{1}{P_1}-\frac{1}{P_2}-\frac{1}{P_4}-\frac{1}{P_5}\bigg)$$

I could plug these values back into the first equation to get $P(A)$, but this method seems extremely cumbersome and I've probably either made a mistake, or I'm missing a way to simplify this. There are ways to go from A to B that are common to both when bridge $3$ is open or closed. Have I overcounted somehow?
There are five bridges and there are 32 states of the bridges.

The probabilities of the 32 states can be easily calculated.

The probability of event $A$ is the sum of those probabilities that belong to states that make it possible to get through.

The conditional probability can be calculated the same way -- listing states...
1.\begin{align*} P(A) &amp;= (1-B_3)(B_1B_4 + B_2B_5) + B_3(B_1B_4 + B_1B_5 + B_2B_4 + B_2B_5)\\
        &amp;= B_1B_4 + B_2B_5 + B_3B_1B_5 + B_3B_2B_4\end{align*}


$$P(A \text{ and } B_3) = B_3(B_1B_4 + B_1B_5 + B_2B_4 + B_2B_5) $$
Answer is equal to $P(A \text{ and }B_3) / P(A).$