Answer:
40% probability that the package was weighed by Employee C
Step-by-step explanation:
Bayes Theorem:
Two events, A and B.
![P(B|A) = \frac{P(B)*P(A|B)}{P(A)}](https://tex.z-dn.net/?f=P%28B%7CA%29%20%3D%20%5Cfrac%7BP%28B%29%2AP%28A%7CB%29%7D%7BP%28A%29%7D)
In which P(B|A) is the probability of B happening when A has happened and P(A|B) is the probability of A happening when B has happened.
In this question:
Event A: Incorrect package.
Event B: Weighed by Employee C.
A supermarket has three employees who equally packages and weighs produce.
This means that ![P(B) = \frac{1}{3}](https://tex.z-dn.net/?f=P%28B%29%20%3D%20%5Cfrac%7B1%7D%7B3%7D)
Employee C records the correct weight 94% of the time
So incorrectly 6% of the time, which means that ![P(A|B) = 0.06](https://tex.z-dn.net/?f=P%28A%7CB%29%20%3D%200.06)
Probability of an incorrect package:
4% of 1/3(A)
5% of 1/3(B)
6% of 1/3(C)
So
![P(A) = \frac{0.04 + 0.05 + 0.06}{3} = 0.05](https://tex.z-dn.net/?f=P%28A%29%20%3D%20%5Cfrac%7B0.04%20%2B%200.05%20%2B%200.06%7D%7B3%7D%20%3D%200.05)
What is the probability that the package was weighed by Employee C?
![P(B|A) = \frac{\frac{1}{3}*0.06}{0.05} = 0.4](https://tex.z-dn.net/?f=P%28B%7CA%29%20%3D%20%5Cfrac%7B%5Cfrac%7B1%7D%7B3%7D%2A0.06%7D%7B0.05%7D%20%3D%200.4)
40% probability that the package was weighed by Employee C