A system has 4 components. We want to find the probability that in some implementation of this system components #1 and #2 fail, and at least one of components #3 and #4 fails as well. (a) Compute this probability under the following two assumptions • component #x fails 25% of the times and component #y fails 30% of the times, where x = 1, 2 and y = 3, 4, • the components fail independently of one another.

Question

Mathematics

Joyce Higgins

25 July, 18:39

A system has 4 components. We want to find the probability that in some implementation of this system components #1 and #2 fail, and at least one of components #3 and #4 fails as well. (a) Compute this probability under the following two assumptions • component #x fails 25% of the times and component #y fails 30% of the times, where x = 1, 2 and y = 3, 4, • the components fail independently of one another.

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Michael Rojas · Answer 1

The probability that components #1 and #2 fail, and at least one of components #3 and #4 fails is 0.031875

Step-by-step explanation:

Let's define the following events,

F1: Component #1 fails

F2: Component #2 fails

F3: Component #3 fails

F4: Component #4 fails

We are looking for the following probability

P[ (F1∩F2) ∩ (F3∪F4) ]

using distributive laws we have

P[ (F1∩F2) ∩ (F3∪F4) ] = P[ (F1∩F2∩F3) ∪ (F1∩F2∩F4) ] = P (F1∩F2∩F3) + P (F1∩F2∩F4) - P (F1∩F2∩F3∩F4) = P (F1) P (F2) P (F3) + P (F1) P (F2) P (F4) - P (F1) P (F2) P (F3) P (F4) = (0.25) (0.25) (0.3) + (0.25) (0.25) (0.3) - (0.25) (0.25) (0.3) (0.3) = 0.031875

by the assumption of independence (the components fail independently) and because P (F1) = P (F2) = 0.25, P (F3) = P (F4) = 0.3