# Stats - Independence (circuits)

1. Feb 4, 2006

### mattmns

Hello, my book has this question, and no examples (very) similar to it, so I am wondering if I did it correct
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The following circuit operates if and only if there is a path of functional devices from left to right. The probability that deach device functions is as shown. Assume that the probability that a device is functional does not depend on whether or not other devices are functional. What is the probability that the circuit operates?

---|----0.9------0.8-------0.7-----|
---|-------------------------------|-----
---|----0.95----0.95------0.95-----|

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So what I did was:

Let T denote the event that the entire top is true, meaning that everything functions on the top.
Let B denote the event that the entire bottom is true.

$P(T \cup B) = 1 - P(T' \cap B')$
$P(T' \cap B') = P(T')P(B')$
$P(T') = 1 - P(T1 \cap T2 \cap T3) = 1 - P(T1)P(T2)P(T3)$ T1 is the first thing on the top being functional (meaning 0.9), T2 being 0.8, etc
$P(B') = 1 - P(B1 \cap B2 \cap B3) = 1 - P(B1)P(B2)P(B3)$
$P(T') = 1 - (0.9)(0.8)(0.7) = 0.496$
$P(B') = 1 - (.95)(.95)(.95) = 0.142625$
$P(T' \cap B') = (0.496)(0.142625) = 0.070742$
$P(T \cup B) = 1 - 0.070742 = 0.929258$

Does that look correct?
Thanks!

Last edited: Feb 4, 2006
2. Mar 30, 2006

$$P(A\cup B) = P(A) + P(B) - P(A\cap B)$$
So $$P(A) = (0.9)(0.8)(0.7).... P(B) = (0.95)(0.95)(0.95).... P(A\cap B) = (0.9)(0.8)(0.7)(0.95)(0.95)(0.95)....$$