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?

pezgirl79u

pezgirl79u

Answered question

2022-10-23

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?

Answer & Explanation

Messiah Trevino

Messiah Trevino

Beginner2022-10-24Added 18 answers

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 A 2 ) = P ( A 1 ) P ( A 2 ) = 0.025 0.034 = 0.00085
since it is given that they are independent. If they were in fact mutually exclusive, then P ( A 1 A 2 ) = P ( ) = 0, which is not the case.
Since the events are not mutually exclusive, then by inclusion-exclusion, we would compute
P ( A 1 A 2 A 3 ) = P ( A 1 ) + P ( A 2 ) + P ( A 3 ) P ( A 1 A 2 ) P ( A 1 A 3 ) P ( A 2 A 3 ) + P ( A 1 A 2 A 3 ) .
Instead, you use the complement,
P ( At least one pump fails ) = 1 P ( No pump fails ) = 1 P ( A ¯ 1 A ¯ 2 A ¯ 3 ) = 1 P ( A ¯ 1 ) P ( A ¯ 2 ) P ( A ¯ 3 ) = 1 ( 1 0.025 ) ( 1 0.034 ) ( 1 0.02 ) = 0.076987.

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