A system conists of two identical pumps, Pump #1 and Pump #2, arranged in parall

Question

A system conists of two identical pumps, Pump #1 and Pump #2, arranged in parallel. If one pump fails, the system will still operate, but the added strain makes it more likely that the remaining pump will fail. For systems designed in this manner and using these pumps, the probability that at least one pump will fail by the end of the pump design life is 0.07, and the probablity that both pumps fail during this time is 0.01. Find the following two probabilities:

[a.] A given pump of this type will fail during its design life.
[b.] The probability that exactly 1 pump will fail during the design life.

Hint from Professor:

The question mentions a particular type of pump being used. This pump has some base rate of failure. Since the system has two pumps, the rate of failure of the whole system is different than the probabilities given.

Part (a) is asking for the probability that a pump will fail, that is: The base rate of failure for an individual pump.

Part (b) is asking for the probability that one of the pumps will fail (it could be either of the pumps) but that the system will still work.

Explanation / Answer

since the pumps are identical , the probability of failure for pump 1 and pump 2 are equal

P(1)= P(2)

Given P( 1 or 2 ) = 0.07 ; P(1 and 2 ) =0.01

using addition theorem of probabilities P(1 or 2 ) = P(1)+P(2) - P(1 and 2)

0.07= 2P(1)-0.01

P(1) = 0.08/2 =0.04; P(2) = 0.04

a) A given pump of this type will fail during its design life is = 0.04
[b.] The probability that exactly 1 pump will fail during the design life.

P( 1 and not 2 ) +( not 1 but 2 ) = (0.04-0.01) + (0.04-0.01) =0.06

Alternatively ; P( exactly one) = P( 1 or 2 ) - P(1 and 2 ) = 0.07-0.01 = 0.06

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