Answer:
The probability that it came from A, given that is defective is 0.362.
Step-by-step explanation:
Define the events:
A: The element comes from A.
B: The element comes from B.
C: The element comes from C.
D: The elemens is defective.
We are given that P(A) = 0.25, P(B) = 0.35, P(C) = 0.4. Recall that since the element comes from only one of the machines, if an element is defective, it comes either from A, B or C. Using the probability axioms, we can calculate that

Recall that given events E,F the conditional probability of E given F is defined as
, from where we deduce that
.
We are given that given that the element is from A, the probability of being defective is 5%. That is P(D|A) =0.05. Using the previous analysis we get that

We are told to calculate P(A|D), then using the formulas we have
