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.
My possible answers are:
.7559
.2441
.8217
.8128
I have tried and asked and no one is getting any of these. Help, please!
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.
To find the probability that the machine works properly, we need to calculate the probability that all four parts, A, B, C, and D, work.
Since the probability of one part working does not depend on the functionality of any of the other parts, we can multiply the probabilities of each part working together to find the overall probability.
P(A and B and C and D) = P(A) * P(B) * P(C) * P(D)
Plugging in the given values:
P(A and B and C and D) = 0.93 * 0.93 * 0.95 * 0.92
Multiplying these values together:
P(A and B and C and D) = 0.75597
Rounding to four decimal places, the probability that the machine works properly is approximately 0.7560.
Therefore, the correct answer from the given options is .7559.
To find the probability that the machine works properly, we can use the concept of independent events. Since all four components must work for the machine to work properly, we need to find the probability that all four components work simultaneously.
Assuming the probabilities of the individual parts working are independent, the probability that all four components work can be calculated by multiplying the probabilities of each component working.
So, the probability that the machine works properly can be calculated as follows:
P(machine works) = P(A) * P(B) * P(C) * P(D)
Given: P(A) = 0.93, P(B) = 0.93, P(C) = 0.95, and P(D) = 0.92
P(machine works) = 0.93 * 0.93 * 0.95 * 0.92
Calculating this expression:
P(machine works) = 0.7866
The answer closest to this value among the options provided is 0.7559. Therefore, the correct answer is 0.7559.
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.