Answer :
Answer:
40% probability that the package was weighed by Employee C
Step-by-step explanation:
Bayes Theorem:
Two events, A and B.
[tex]P(B|A) = \frac{P(B)*P(A|B)}{P(A)}[/tex]
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 [tex]P(B) = \frac{1}{3}[/tex]
Employee C records the correct weight 94% of the time
So incorrectly 6% of the time, which means that [tex]P(A|B) = 0.06[/tex]
Probability of an incorrect package:
4% of 1/3(A)
5% of 1/3(B)
6% of 1/3(C)
So
[tex]P(A) = \frac{0.04 + 0.05 + 0.06}{3} = 0.05[/tex]
What is the probability that the package was weighed by Employee C?
[tex]P(B|A) = \frac{\frac{1}{3}*0.06}{0.05} = 0.4[/tex]
40% probability that the package was weighed by Employee C