Question

4. Two technicians regularly make repairs when breakdowns occur on an automated production line. The first technician, who services 40% of the breakdowns, has 5% chance of making incomplete repair. The second technician, who services 60% of the breakdowns, has a 3% chance of making an incomplete repair. Given that there is a problem with the production line due to an incomplete repair, what is the probability that this initial repair was made by the first technician

Answer 1

Answer:

52.63% probability that this initial repair was made by the first technician

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: Error due to an incomplete repair.

Event B: Initial repair made by the first technican.

The first technician, who services 40% of the breakdowns, has 5% chance of making incomplete repair.

This means that $P(B) = 0.4, P(A|B) = 0.05$

Probability of an incomplete repair:

5% of 40%(first technican)

3% of 60%(second technican).

Then

$P(A) = 0.05*0.4 + 0.03*0.6 = 0.038$

Given that there is a problem with the production line due to an incomplete repair, what is the probability that this initial repair was made by the first technician

$P(B|A) = \frac{0.4*0.05}{0.038} = 0.5263$

52.63% probability that this initial repair was made by the first technician