The Problem: What is the probability each circuit will be connected? The probability each switch will be connected is $p\!$ and $P(A)=\frac{2}{3}$ and $P(B)=\frac{1}{3}$.

The probablity of the top path (path $A\!$) being connected is $1-(1-p)^2\!$, which is found by finding the probability the path won't be connected and subtracting it from 1. The probability of the bottom path (path $B\!$) is obviously $p^2\!$. Thus

$P(OC)=\frac{2}{3}[1-(1-p)^2]+\frac{1}{3}p^2$

For this design note it is simply the previous problem with another option added in the middle. Thus we can say

$P(OC)=P(OC|MC)P(MC)+P(OC|MNC)P(MNC)\!$ where

$OC = \!$ Overall Connected
$MC = \!$ Middle Connected
$MNC = \!$ Middle Not Connected

and

$P(OC|MC)\!$ is $1\!$ (if the middle is connected then the entire circuit is!)
$P(MC)\!$ is just $p\!$ (probability of the middle switch being connected)
$P(OC|MNC)\!$ is simply the answer from the first part
$P(MNC)\!$ is one minus the probability the middle is connected, so $1-p\!$

Thus:
$P(OC)=1p + (1-p)[\frac{2}{3}[1-(1-p)^2]+\frac{1}{3}p^2]\!$

## Alumni Liaison

has a message for current ECE438 students.

Sean Hu, ECE PhD 2009