Find the probability that the entire system will operate without failure

[toothpaste666]

Homework Statement

Suppose components A, B1 and B2 operate independently, in a system below, and the probabilities that anyone of the components will operate for 100 hours without failure are P(A) = 0.8,P(B1) = P(B2) = 0.9. The system works if A works and either B1 or B2 works. Find the probability that the entire system will operate without failure for 100 hours.

The Attempt at a Solution

the probability that the system runs for 100 hours without failure is equal to
P(A) and (P(B1) or P(B2))

for any events P(A) or P(B) = P(A) + P(B) - P(A and B)

so P(B1) or P(B2) = P(B1) + P(B2) - P(B1 and B2)

since B1 and B2 are independent events, P(B1 and B2) = P(B1)P(B2)

so P(B1) or P(B2) = P(B1) + P(B2) - P(B1)P(B2) = .9 + .9 - (.9)(.9) = .99

so the probability that the system runs without failure for 100 hours is
P(A) and (.99)
since both of these are independent events
P(A) and (.99) = P(A)(.99) = (.8)(.99) = .792

is this correct?

 

[RUber]

Yep.

 

[toothpaste666]

Thanks again! :D

 

[RUber]

Just a note for future problems like this, Often when only one of a set needs to be working in order for the system to work, it can be written as P(at least one works ) = 1- P(none work). Your method was correct, but I am sure you can imagine with there were B1, B2, ..., B16 redundant elements, you would not want to work out all the cases of 1 or more elements working. And, even for two elements, the math is about 1 step shorter.

 

[toothpaste666]

I see what you are saying. so (P(B1) or P(B2)) = 1 - P(!B1 and !B2) = 1 - P(!B1)P(!B2) = 1 - (.1)(.1) = .99
yeah that is much easier. thank you!