Multiple choice: Probability

Question

A particular system is composed of three machine. A, B, and C. The entire system will work as long as any one (or more than one of the three machines works. (The system is WITH backup).

P(A works) = 0.60

P(B works) = 0.70

P(C works) = 0.85

Round to the fourth digit beyond the decimal point (if it goes that far)

P(the entire system works) =

Answer 1

To solve this problem, we will need to understand the concept of the complement of an event.

The complement of an event is the probability that an event DOESN'T occur.

An event and its complement are the only two possible outcomes, and since the sum of the probabilities of all possible outcomes must equal one (This is one of the fundamental rules or axioms of probability.) :

P(A) + P(A complement) = 1

P(A complement ) = 1- P(A)

The probability that a machine is NOT working is simply the complement of the probability that a machine is working.

P(A does not work)) = 1- P(A works) = 1- 0.60 = 0.40

P(B does not work) = 1 - P (B works) = 1- 0.70 = 0.30

P(C does not work) = 1- P (C works) = 1 - 0.85 = 0.15

The system works so long as at least one of the machines is working.

If all three machines break down, the system will fail.

P(System Fails) = P(A does not work) × P(B does not work) × P(C does not work)

= 0.40 × 0.30 × 0.15 = 0.018

Hang on a minute! We are not done. We are asked for the probability that the system works. Well the probability that the system works is the complement of the probability that the system fails.

P(System Works)= 1- P(System Fails) = 1 - 0.018 = 0.982

Probability is Super Fun!

Answer 2

Hello Louie!

We must find the probability of neither of them works.

P(A)^c = 1 - P(A) = 0.40

P(B)^c = 1 - P(B) = 0.30

P(C)^c = 1 - P(C) = 0.15

P^c = P(A)^c * P(B)^c * P(C)^c = 0.40*0.30*0.15 = 0.018

P = 1 - P^c = 0.982