Q. 3.66 - A First Course in Probability

Question

The probability of the closing of the $i\text{ th }$ relay in the circuits shown in Figure $3.4$ is given by $p_i,i=1,2,3,4,5$. If all relays function independently, what is the probability that a current flows between $A$ and  for the respective circuits.

             

Step-by-Step Solution

Answer

The probability that a current flows between $A$ and $B$ for the respective circuits are

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

$\text{ b) }{p}_1{p}_4+p_2{p}_5+p_3\left(p_1{p}_5+p_2{p}_4\right)$ $-\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$

  

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

Step 1: Given

Electric circuit from $A$ to $B$

$5$ independent switches

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

Sketches for a) and b)

     

Step 2: Inclusion and Exclusion Formula

we are able to see how the seems as via either shift $1,2,5$ or through $3,4,5$ . The addition and exemption method are going to be utilized in the highest   row, while a freedom method is is employed for the last.

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

$=P\left(C_1{C}_2{C}_5\right)+P\left(C_3{C}_4{C}_5\right)-P\left(C_1{C}_2{C}_3{C}_4{C}_5\right)$    $=P\left(C_1\right)P\left(C_2\right)P\left(C_5\right)+P\left(C_3\right)P\left(C_4\right)P\left(C_5\right)-P\left(C_1\right)P\left(C_2\right)P\left(C_3\right)P\left(C_4\right)P\left(C_5\right)$

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

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

Step 3: Closed the present

If $1$ and $4$ are blocked, or $2$ and $5$, this  can flow.

When button 3 is locked, power can move valves $1,3,5$ or $2,3,4$ too.

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

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

$\begin{array}{r}=P\left(C_3^c\right)\left[P\left(C_1{C}_4\right)+P\left(C_2{C}_5\right)-P\left(C_1{C}_2{C}_4{C}_5\right)\right]+P\left(C_3\right)\left[P\left(C_1{C}_4\right)+P\left(C_2{C}_5\right)+\right.\\ P\left(C_1{C}_5\right)+P\left(C_2{C}_4\right)-P\left(C_1{G}_4{C}_2{C}_5\right)-P\left(C_1{C}_4{C}_5\right)-P\left(C_1{C}_2{C}_4\right)-P\left(C_2{C}_5{C}_1\right)-\\ P\left(C_2{C}_5{C}_4\right)-\underset{\_}{P\left(C_4{C}_2{C}_4{C}_5\right)}+A_{1}P\left(C_1{C}_2{C}_5{C}_4\right)-\underset{\_}{\left.P\left(C_1{C}_2{C}_5{C}_4\right)\right]}\end{array}$

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

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

$\begin{array}{r}=\left[P\left(C_3^c\right)+P\left(C_3\right)\right]\left[P\left(C_1{C}_4\right)+P\left(C_2{C}_5\right)\right]-P\left(C_3^c\right)P\left(C_1{C}_2{C}_4{C}_5\right)+\\ P\left(C_3\right)\left[P\left(C_1{C}_5\right)+P\left(C_2{C}_4\right)-P\left(C_1{C}_2{C}_4\right)-P\left(C_1{C}_2{C}_5\right)-\right.\\ \left.P\left(C_1{C}_4{C}_5\right)-P\left(C_2{C}_4{C}_5\right)+P\left(C_1{C}_2{C}_4{C}_5\right)\right]\end{array}$