Posterior Probability: Solving Questions on Machine Adjustment

[anjunabeats]

Have a question that I'm having some difficulties with, wondering if I'm on the right track:

A machine has a 0.9 probability that it is adjusted correctly and a 0.1 probability that it is not adjusted correctly.
When it is properly adjusted, it produces good items 1/2 of the time and bad items the other half of the time.
When it is incorrectly adjusted, it produces good items 1/4 of the time and bad items 3/4 of the time.

a) suppose that five items are produced, 4 of which are good items, 1 of which is a bad item.
What is the P that the machine is correctly adjusted?

I get 96/97 = 98.97%

Workings:

0.9 x 5C4 x 0.5^5
divided by:
(0.9 x 5C4 x 0.5^5) + (0.1 x 5C4 x 0.25^4 x 0.75)


b) Suppose one additional item is produced by the machine at the same time as the other 5 items and is found to be of medium quality. What is the new posterior probability that the machine was adjusted properly?

I'm not sure exactly what is meant by this but I used the same working as the above but replaced with 6C4 instead and got 64/65 = 98.46%

Am I doing something wrong here? Probability is not my strong point!
Thanks for your help and replies.

 

[Valkarie]

Yes, you are on the right track. The way to solve this question is to use Bayes' theorem, which states that the posterior probability of an event A given another event B (P(A|B)) is equal to the probability of B given A (P(B|A)) multiplied by the prior probability of A (P(A)) divided by the probability of B (P(B)). In this case, P(A) = 0.9, P(B) = 0.5^5 + 0.25^4 x 0.75, P(B|A) = 0.5^6, and P(A|B) = ?. So, the posterior probability of the machine being adjusted properly given that one item was of medium quality is P(A|B) = (0.9 x 0.5^6) / (0.5^5 + 0.25^4 x 0.75) = 64/65 = 98.46%.