Question

The probability of the closing of the ith relay in the circuits shown in Figure 3.4 is given by pi,$i=1,2,3,4,5$. If all relays function independently, what is the probability that a current flows between A and B for the respective circuits?

Short answer

a) The probability is$\left(p_1{p}_2+p_3{p}_4-p_1{p}_2{p}_3{p}_4\right)p_5$

b)The probability is

$$\begin{gathered} p_1{p}_4+p_2{p}_5+p_3\left(p_1{p}_5+p_2{p}_4\right) \\ -\left(p_1{p}_2{p}_3{p}_4+p_1{p}_2{p}_3{p}_5+p_1{p}_2{p}_4{p}_5+p_1{p}_3{p}_4{p}_5+p_2{p}_3{p}_4{p}_5\right)+2p_1{p}_2{p}_3{p}_4{p}_5 \end{gathered}$$

Step-by-step solution

Given Information(Part a)

Electrical circuit from $A$to $B$

$5$independent switches

$C_i$- event that switch $i$is closed,

$P\left(C_i\right)=p_i,i=1,2,3,4,5$

$P\left(E\right)$, the probability that the current flows.

Explanation (Part a)

We see that the current flows either through switches $1,2,5$ or through $3,4,5$. The first row uses the Inclusion and Exclusion formula and the second the independence

$P\left(E\right)=P\left(C_1{C}_2{C}_5\cup C_3{C}_4{C}_5\right)$

$=P\left(C_1{C}_2{C}_5\right)+P\left(C_3{C}_4{C}_5\right)-P\left(C_1{C}_2{C}_3{C}_4{C}_5\right)$

$=P\left(C_1\right)P\left(C_2\right)P\left(C_5\right)+P\left(C_3\right)P\left(C_4\right)P\left(C_5\right)-P\left(C_1\right)P\left(C_2\right)P\left(C_3\right)P\left(C_4\right)P\left(C_5\right)$

$=p_1{p}_2{p}_5+p_3{p}_4{p}_5-p_1{p}_2{p}_3{p}_4{p}_5$

$=\left(p_1{p}_2+p_3{p}_4-p_1{p}_2{p}_3{p}_4\right)p_5$

Final Answer (Part a)

$\left(p_1{p}_2+p_3{p}_4-p_1{p}_2{p}_3{p}_4\right)p_5$

Given Information (Part b)

Electrical circuit from $A$to $B$

$5$independent switches

$C_i$- event that switch $i$is closed,

$P\left(C_i\right)=p_i,i=1,2,3,4,5$.

Explanation (Part b)

The current flows if $1$ and $4$ are closed or $2$ and $5$ are closed.

If the switch $3$ is closed the current can flow also through switches $1,3,5$ or through $2,3,4$.

$P\left(E\right)=P\left(C_1{C}_4\cup C_2{C}_5\cup C_3{C}_1{C}_5\cup C_3{C}_2{C}_4\right)$

$=P\left[C_3^c\left(C_1{C}_4\cup C_2{C}_5\right)\cup C_3\left(C_1{C}_4\cup C_2{C}_5\cup C_1{C}_5\cup C_2{C}_4\right)\right]$

$=P\left(C_3^c\right)P\left(C_1{C}_4\cup C_2{C}_5\right)+P\left(C_3\right)P\left(C_1{C}_4\cup C_2{C}_5\cup C_1{C}_5\cup C_2{C}_4\right)$

$=p_1{p}_4+p_2{p}_5-p_1{p}_2{p}_4{p}_5+p_1{p}_2{p}_3{p}_4{p}_5+p_3\left(p_1{p}_5+p_2{p}_4-\right.\left.p_1{p}_2{p}_4-p_1{p}_2{p}_5-p_1{p}_4{p}_5-p_2{p}_4{p}_5\right)+p_1{p}_2{p}_3{p}_4{p}_5$

role="math" localid="1647859751713" $=p_1{p}_4+p_2{p}_5+p_3\left(p_1{p}_5+p_2{p}_4\right)-\left(p_1{p}_2{p}_3{p}_4+p_1{p}_2{p}_3{p}_5+p_1{p}_2{p}_4{p}_5+p_1{p}_3{p}_4{p}_5+p_2{p}_3{p}_4{p}_5\right)+2p_1{p}_2{p}_3{p}_4{p}_5$

Final Answer (Part b)

The probability is

$p_1{p}_4+p_2{p}_5+p_3\left(p_1{p}_5+p_2{p}_4\right)-\left(p_1{p}_2{p}_3{p}_4+p_1{p}_2{p}_3{p}_5+p_1{p}_2{p}_4{p}_5+p_1{p}_3{p}_4{p}_5+p_2{p}_3{p}_4{p}_5\right)+2p_1{p}_2{p}_3{p}_4{p}_5$