91影视

$E_i$- The current can flow through switch $i$.

$C$- Event that the current flows from $A$to $B$.

$P_i$- Probability that a certain switch is closed.

The switches are independent.

It is given that we have to start with total probability conditioning on $E_1$.

$P\left(C\right)=P\left(C鈭�E_1\right)P\left(E_1\right)+P\left(C鈭�E_1^c\right)P\left(E_1^c\right)$

Now regard conditional probability $P\left(路鈭�E_1\right)$and $P\left(路鈭�E_1^c\right)$as a probability and condition on $E_3$:

$$\begin{gathered} P\left(C\right)=\left[P\left(C鈭�E_3{E}_1\right)P\left(E_3鈭�E_1\right)+P\left(C鈭�E_3^c{E}_1\right)P\left(E_3^c鈭�E_1\right)\right]P\left(E_1\right)+ \\ \left[P\left(C鈭�E_3{E}_1^c\right)P\left(E_3鈭�E_1^c\right)+P\left(C鈭�E_3^c{E}_1^c\right)P\left(E_3^c鈭�E_1^c\right)\right]P\left(E_1^c\right) \end{gathered}$$

Now, with knowledge about switches 1 and 3 , show $C$as combination of $E_2,E_4,E_5$, that is:

$P\left(C鈭�E_3{E}_1\right)=P\left(E_4鈭�E_5鈭�E_3{E}_1\right){\stackrel{铃�}{=}}^{\text{independence}}P\left(E_4鈭�E_5\right)=P_4+P_5-P_4{P}_5$

$P\left(C鈭�E_3^c{E}_1\right)=P\left(E_4鈭�E_3^c{E}_1\right)=P\left(E_4\right)=P_4$

$P\left(C鈭�E_3{E}_1^c\right)=P\left(E_2鈭�\left(E_4鈭�E_5\right)鈭�E_3{E}_1\right)=P\left(E_2鈭�\left(E_4鈭�E_5\right)\right)=P_2\left(P_4+P_5-P_4{P}_5\right)$

$P\left(C鈭�E_3^c{E}_1^c\right)=P\left(E_2{E}_5鈭�E_3^c{E}_1^c\right)=P\left(E_2{E}_5\right)=P_2{P}_5$

This is because we can reduce $C$ to stricter combinations of $E_2,E_4,E_5$, given the knowledge about $E_1,E_3$

Use independence, and probabilities above to transform the last formula of total probability:

$$\begin{gathered} P\left(C\right)=\left[\left(P_4+P_5-P_4{P}_5\right)P_3+P_4\left(1-P_3\right)\right]P_1+ \\ \left[\left(P_2\left(P_4+P_5-P_4{P}_5\right)\right)P_3+\left(P_2{P}_5\right)\left(1-P_3\right)\right]\left(1-P_1\right) \end{gathered}$$

The probability that a current flows $A$from $B$is

$$\begin{gathered} P\left(C\right)=\left[\left(P_4+P_5-P_4{P}_5\right)P_3+P_4\left(1-P_3\right)\right]P_1+ \\ \left[\left(P_2\left(P_4+P_5-P_4{P}_5\right)\right)P_3+\left(P_2{P}_5\right)\left(1-P_3\right)\right]\left(1-P_1\right) \end{gathered}$$

$E_i$ - the current can flow through switch $i$.

$C$- event that the current flows from $A$ to $B$.

$P_i$ - probability that a certain switch is closed.

Start with the definition:

$P\left(C\right)$ is known from a) and:

And $C{E}_3^c=E_1{E}_4鈭�E_2{E}_5$thus

$P\left(C{E}_3^c\right)=P\left(E_1{E}_4鈭�E_2{E}_5\right)$

$=P\left(E_1{E}_4\right)+P\left(E_2{E}_5\right)-P\left(E_1{E}_4{E}_2{E}_5\right)$

$=P_1{P}_4+P_2{P}_5-P_1{P}_2{P}_4{P}_5$

$$\begin{gathered} P\left(E_{3}C\right)=\left[\left(P_4+P_5-P_4{P}_5\right)P_3+P_4\left(1-P_3\right)\right]P_1+\left[\left(P_2\left(P_4+P_5-P_4{P}_5\right)\right)P_3\right. \\ \left.+\left(P_2{P}_5\right)\left(1-P_3\right)\right]\left(1-P_1\right)-P_1{P}_4-P_2{P}_5+P_1{P}_2{P}_4{P}_5 \end{gathered}$$

$$\begin{gathered} \left.=\left[\left(P_4+P_5-P_4{P}_5\right)P_3-P_4{P}_3\right)\right]P_1+\left[\left(P_2\left(P_4+P_5-P_4{P}_5\right)\right)P_3\right. \\ \left.+\left(P_2{P}_5\right)\left(1-P_3\right)\right]\left(1-P_1\right)-P_2{P}_5+P_1{P}_2{P}_4{P}_5 \end{gathered}$$

And now substitute this into the definition:

$P\left(E_3\right|C)=\frac{\left.\left[\left(P_4+P_5-P_4{P}_5\right)P_3-P_4{P}_3\right)\right]P_1+\left[\left(P_2\left(P_4+P_5-P_4{P}_5\right)\right)P_3+\left(P_2{P}_5\right)\left(1-P_3\right)\right]\left(1-P_1\right)-P_2{P}_5+P_1{P}_2{P}_4{P}_5}{\left[\left(P_4+P_5-P_4{P}_5\right)P_3+P_4\left(1-P_3\right)\right]P_1+\left[\left(P_2\left(P_4+P_5-P_4{P}_5\right)\right)P_3+\left(P_2{P}_5\right)\left(1-P_3\right)\right]\left(1-P_1\right)}$

The conditional probability is$\frac{\left.\left[\left(P_4+P_5-P_4{P}_5\right)P_3-P_4{P}_3\right)\right]P_1+\left[\left(P_2\left(P_4+P_5-P_4{P}_5\right)\right)P_3+\left(P_2{P}_5\right)\left(1-P_3\right)\right]\left(1-P_1\right)-P_2{P}_5+P_1{P}_2{P}_4{P}_5}{\left[\left(P_4+P_5-P_4{P}_5\right)P_3+P_4\left(1-P_3\right)\right]P_1+\left[\left(P_2\left(P_4+P_5-P_4{P}_5\right)\right)P_3+\left(P_2{P}_5\right)\left(1-P_3\right)\right]\left(1-P_1\right)}$