A machine has four components, A, B, C, and D, set up in such a manner that all four parts must work for the machine to work properly. Assume the probability of one part working does not depend on the functionality of any of the other parts. Also assume that the probabilities of the individual parts working are P(A) = P(B) = 0.93, P(C) = 0.95, and P(D) = 0.92. Find the probability that the machine works properly.

Question

My possible answers are:
.7559
.2441
.8217
.8128

I have tried and asked and no one is getting any of these. Help, please!

Answer 1

My understanding of your question means that you have a fault tree with 'machine unavailable' as the top event and the Boolean algebra expression for the top event being

A OR B OR C OR D

because failure of any one of the four components results in machine failure.

As an approximation (because it will be missing the AB components and therefore slightly too high - the probability of A OR B is [P(A)+P(B)-P(A)*P(B)])

is 0.07+0.07+0.05+0.08 = 0.27, which is slightly high so the unavailability is likely to be the 0.2441 value, hence the probability that it is working is

1-0.2441 = 0.7559.

You will need to workout an exact value for your answer and the best way to do this is to sequentially use the [P(A)+P(B)-P(A)*P(B)] expression. What I mean is combine A OR B, then combine the result of this with C and so on.

Answer 2

Just to confirm.

If you go through the maths steps as I indicated I got

0.2439854, so allowing for the occasion rounding error this is close to 0.2441.