Question

g A supermarket has three employees who equally packages and weighs produce. Employee A records the correct weight 96% of the time. Employees B and C record the correct weight 95% and 94% of the time, respectively. One customer complains about the incorrect weight recorded on a package he has purchased. What is the probability that the package was weighed by Employee C?

Answer 1

Answer:

40% probability that the package was weighed by Employee C

Step-by-step explanation:

Bayes Theorem:

Two events, A and B.

$P(B|A) = \frac{P(B)*P(A|B)}{P(A)}$

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 $P(B) = \frac{1}{3}$

Employee C records the correct weight 94% of the time

So incorrectly 6% of the time, which means that $P(A|B) = 0.06$

Probability of an incorrect package:

4% of 1/3(A)

5% of 1/3(B)

6% of 1/3(C)

So

$P(A) = \frac{0.04 + 0.05 + 0.06}{3} = 0.05$

What is the probability that the package was weighed by Employee C?

$P(B|A) = \frac{\frac{1}{3}*0.06}{0.05} = 0.4$

40% probability that the package was weighed by Employee C