Question

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.

Answer 1

Answer:

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