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
![P(D) = P(A\cap D) + P(B\cap D) + P(C\cap D)](https://tex.z-dn.net/?f=P%28D%29%20%3D%20P%28A%5Ccap%20D%29%20%2B%20P%28B%5Ccap%20D%29%20%2B%20P%28C%5Ccap%20D%29)
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
![P(D) = P(A)P(D|A)+P(B) P(D|B) + P(C)P(D|C) = 0.05\cdot 0.25+0.04\cdot 0.35+0.02\cdot 0.4 = 0.0345](https://tex.z-dn.net/?f=%20P%28D%29%20%3D%20P%28A%29P%28D%7CA%29%2BP%28B%29%20P%28D%7CB%29%20%2B%20P%28C%29P%28D%7CC%29%20%3D%200.05%5Ccdot%200.25%2B0.04%5Ccdot%200.35%2B0.02%5Ccdot%200.4%20%3D%200.0345)
We are told to calculate P(A|D), then using the formulas we have
![P(A|D) = \frac{P(A\cap D)}{P(D)}= \frac{P(D|A)P(A)}{P(D)}= \frac{0.05\cdot 0.25}{0.0345}= 0.36231884](https://tex.z-dn.net/?f=%20P%28A%7CD%29%20%3D%20%5Cfrac%7BP%28A%5Ccap%20D%29%7D%7BP%28D%29%7D%3D%20%5Cfrac%7BP%28D%7CA%29P%28A%29%7D%7BP%28D%29%7D%3D%20%5Cfrac%7B0.05%5Ccdot%200.25%7D%7B0.0345%7D%3D%200.36231884)