Conditional Probability and Independent Events

[djhembree]

Does anyone know where to start on a problem like this. I look at the formula in the book but it is not helping me.

A parallel system functions whenever at least one of its components works. Consider a parallel system of n components, and consider that each component independently works with probability ½. Find the conditional probability that component 1 works, given that the system is functioning.

 

[soroban]

Hello, djhembree!

A parallel system functions whenever at least one of its components works.
Consider a parallel system of \(\displaystyle n\) components,
where each component independently works with probability 1/2.

Find the conditional probability that component 1 works, given that the system is functioning.

\(\displaystyle \text{Bayes' Theorem: }\;P(\text{unit 1 works }|\text{ system works}) \;=\;\frac{P(\text{unit 1 works }\wedge\text{ system works})}{P(\text{system works})}\) .[1]


\(\displaystyle \text{Numerator: }\:\text{unit 1 works }and\text{ the system works}\)

. . . . . \(\displaystyle P(\text{unit 1 works}) \:=\:\tfrac{1}{2}\)

. . Here's a revelation: If unit 1 works, then the system will work!
. . . . . \(\displaystyle P(\text{system works}) \:=\:100\% \:=\:1\)

. . \(\displaystyle \text{The numerator is: }\:\tfrac{1}{2}\cdot1 \:=\:\tfrac{1}{2}\) .[2]


\(\displaystyle \text{Denominator: }\,\text{system works}\)

. . \(\displaystyle \text{The system does }not\text{ work if all }n\text{ units fail.}\)

. . . . . \(\displaystyle P(\text{system does }not\text{ work}) \;=\;\left(\frac{1}{2}\right)^n \:=\:\frac{1}{2^n}\)

. . \(\displaystyle \text{Hence: }\(\text{ system works}) \:=\:1 - \frac{1}{2^n} \;=\;\frac{2^n-1}{2^n}\) .[3]


Substitute [2] and [3] into [1]:

. . \(\displaystyle P(\text{unit 1 works }|\text{ system works}) \;=\;\frac{\frac{1}{2}}{\frac{2^n-1}{2^n}} \;=\;\frac{2^{n-1}}{2^n-1}\)