Question

In Problem 3.66a, find the conditional probability that relays $1$and $2$are both closed given that a current flows from $A$ to $B$.

Short answer

The conditional probability is$\frac{p_1{p}_2}{p_1{p}_2+p_3{p}_4-p_1{p}_2{p}_3{p}_4}$.

Step-by-step solution

Given Information

Electrical circuit from$A$to $B$

$5$independent switches

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

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

$\text{Compute}P\left(C_1{C}_2\mid E\right)\text{, the probability that switches}1\text{and}2\text{are both closed given that the current flows}$

Explanation

$P\left(E\right)$

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 row independence

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

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

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

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

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

Explanation

$P\left(C_1{C}_2\cap E\right)$

The current flows, and switches $1$and $2$are closed if and only if $C_1{C}_2{C}_5$occurs.

$P\left(C_1{C}_{2}E\right)=P\left[C_1{C}_2\left(C_1{C}_2{C}_5\cup C_3{C}_4{C}_5\right)\right]$

$=P\left[C_1{C}_2\left(C_1{C}_2\cup C_3{C}_4\right)C_5\right]$

$=P\left(C_1{C}_2{C}_5\right)$

$=p_1{p}_2{p}_5$

Apply the condition of probability,

$P\left(C_1{C}_2\mid E\right)=\frac{P\left(C_1{C}_{2}E\right)}{P\left(E\right)}$

To obtain result,

$P\left(C_1{C}_2\mid E\right)=\frac{p_1{p}_2{p}_5}{\left(p_1{p}_2+p_3{p}_4-p_1{p}_2{p}_3{p}_4\right)p_5}=\frac{p_1{p}_2}{p_1{p}_2+p_3{p}_4-p_1{p}_2{p}_3{p}_4}$

Final Answer

The conditional probability is$\frac{p_1{p}_2}{p_1{p}_2+p_3{p}_4-p_1{p}_2{p}_3{p}_4}$.