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There's a problem:

A city water supply system involved three pumps, the failure of any one of which crashes the system. The probabilities of failure for each pump in a given year are .025, .034, .02, respectively. Assuming the pumps operate independently of each other, what is the probability that the system does crash during the year?

The answer was

1 - (1 - 0.025)(1 - 0.034)(1- 0.02)

I didn't understand that, the answer said that you must first find the probability of all of them not crashing then subtracting that from 1.

Why can't we just add up the probability since these events seem mutually exclusive. Can we not do that if the question suggests independence as it does here?

I don't want you to do any homework for me but just explain the reasoning. That's all I'm asking for.

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  • $\begingroup$ Nontrivial Independent are Never mutually exclusive $\endgroup$
    – JMoravitz
    Apr 24, 2017 at 2:44
  • $\begingroup$ @JMoravitz What do you mean Nontrivial? So if independence is posted then it can't be mutually exclusive? $\endgroup$
    – Asker123
    Apr 24, 2017 at 2:45
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    $\begingroup$ By nontrivial i mean has probability greater than zero. Precisely. Mutually exclusive implies that their intersection is empty. If $A$ and $B$ are mutually exclusive then, $P(A\cap B)=P(\emptyset)=0$. On the other hand if $P(A)>0$ and $P(B)>0$ and $A$ and $B$ are independent, then $P(A\cap B)=P(A)P(B)>0$ and $P(A\cap B)\neq 0$ $\endgroup$
    – JMoravitz
    Apr 24, 2017 at 2:46
  • $\begingroup$ @JMoravitz Thanks bro, you just cleared up something. Because it says independence then an intersection exists so we cannot add them. Am I right? You da man! :D $\endgroup$
    – Asker123
    Apr 24, 2017 at 2:48

2 Answers 2

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Let $A_i$ be the event that pump $i$ fails, with $P(A_1) = 0.025, P(A_2) = 0.034, P( A_3) = 0.02$.

Independence does not imply mutually exclusive. Notice for example that $$P(A_1 \cap A_2) = P(A_1)P(A_2) = 0.025\cdot0.034 = 0.00085$$ since it is given that they are independent. If they were in fact mutually exclusive, then $P(A_1\cap A_2) = P(\varnothing) = 0$, which is not the case.

Since the events are not mutually exclusive, then by inclusion-exclusion, we would compute \begin{align*} P(A_1\cup A_2\cup A_3) &= P( A_1)+P(A_2)+P(A_3)\\ &\quad -P(A_1\cap A_2)-P(A_1\cap A_3)-P(A_2\cap A_3)\\ &\quad +P(A_1\cap A_2\cap A_3). \end{align*}

Instead, you use the complement, \begin{align*} P(\text{At least one pump fails})&=1-P(\text{No pump fails}) \\ &= 1-P(\bar A_1\cap \bar A_2\cap \bar A_3)\\ &=1- P(\bar A_1)P(\bar A_2)P(\bar A_3)\\ &= 1 - (1 - 0.025)(1 - 0.034)(1- 0.02) \\ &=0.076987. \end{align*}

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The answer is simply saying that Such a probability is only when : 1-P(Every pump crashes). Hope it solves your doubt?

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  • $\begingroup$ Why can't we add up the probabilities though? $\endgroup$
    – Asker123
    Apr 24, 2017 at 2:46
  • $\begingroup$ @Asker123 You cannot. Because they are independent events. and for such events, $P(A \cap B)=P(A).P(B)$ $\endgroup$ Apr 24, 2017 at 2:58

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