Q.3.16 - A First Course in Probability

Question

(a) In Problem 3.66b, find the probability that a current flows from $A$ to $B$, by conditioning on whether relay 1 closes.

(b) Find the conditional probability that relay $3$ is closed given that a current flows from $A$ to $B$ .

Step-by-Step Solution

Answer

a). The probability that a current flows $A$ from $B$ is $$\begin{gathered} P\left(C\right)=\left[\left(P_4+P_5-P_4{P}_5\right)P_3+P_4\left(1-P_3\right)\right]P_1+ \\ \left[\left(P_2\left(P_4+P_5-P_4{P}_5\right)\right)P_3+\left(P_2{P}_5\right)\left(1-P_3\right)\right]\left(1-P_1\right) \end{gathered}$$

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

Step 1: Given Information (Part a)

$E_i$ - 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.

Step 2: Explanation (Part a)

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

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

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

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

Step 3: Explanation (Part a)

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

$P\left(C\mid E_3{E}_1\right)=P\left(E_4\cup E_5\mid E_3{E}_1\right){\overbrace{=}}^{\text{independence }}P\left(E_4\cup E_5\right)=P_4+P_5-P_4{P}_5$

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

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

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

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:

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

Step 4: Final Answer (Part a)

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

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

Given Information (Part b)

$E_i$ - 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.

Step 6: Explanation (Part b)

Start with the definition:

$P\left(E_3\mid C\right)=\frac{P\left(E_{3}C\right)}{P\left(C\right)}$

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

$P\left(E_{3}C\right)=P\left(C\right)-P\left(C{E}_3^c\right)$

And $C{E}_3^c=E_1{E}_4\cup E_2{E}_5$ thus

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

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

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

Step 7: Explanation (Part b)

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

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

And now substitute this into the definition:

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

Final Answer (Part b)

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