Asystem 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.

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

check above link.

graphically?   well, you graph them both, the solutions is where they both intersect each other, since there are two solutions, meaning they will meet twice, or namely the linear will run into the quadratic curve twice, and where they meet, those x,y coordinates are the solutions to the system.