Consider the system of components connected as in the accompanying picture. Components 1 and 2 are connected in parallel, so that sub-system works if and only if either 1 or 2 works; since 3 and 4 are connected in series, that sub-system works if and only if both 3 and 4 work. If components work independently of one another and P(component works) = 0.80, calculate P(system works).

P(system works) = 0.6144

Step-by-step explanation:

Each compoent has an 80% = 0.8 probability of working correctly.

1 and 2

We need at least one of them working.

Either neither works, or at least one works.

Each one has a 1 - 0.8 = 0.2 probability of not working.

0.2*0.2 = 0.04

0.04 of neither working.

1 - 0.04 = 0.96 probability of at least one working

3 and 4

We need both to work, each with a 0.8 probability.

Calculate P(system works).

At least one of the first two, with 0.96 probability

3 and 4, each with 0.8 probability.

0.96*0.8*0.8 = 0.6144

So

P(system works) = 0.6144

yes

step-by-step explanation:

so of course you want to start by plugging the 5 in which goes to 7(5)+3(5)=50 because 7*5=35 and 3*5=15 and then add them together, then plug in the 5 to the other equation as well 9(5)+5=50 because 9*5= 45 and then add 5 and so yes they are equivalent.