# Probability involving independence and mutually exclusive events

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?

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.

• Nontrivial Independent are Never mutually exclusive Apr 24, 2017 at 2:44
• @JMoravitz What do you mean Nontrivial? So if independence is posted then it can't be mutually exclusive? Apr 24, 2017 at 2:45
• 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 $P(A\cap B)\neq 0$ Apr 24, 2017 at 2:46
• @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 Apr 24, 2017 at 2:48

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.
• @Asker123 You cannot. Because they are independent events. and for such events, $P(A \cap B)=P(A).P(B)$ Apr 24, 2017 at 2:58