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.
[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: 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 [tex]P(B) = 0.4, P(A|B) = 0.05[/tex]
Probability of an incomplete repair:
5% of 40%(first technican)
3% of 60%(second technican).
Then
[tex]P(A) = 0.05*0.4 + 0.03*0.6 = 0.038[/tex]
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
[tex]P(B|A) = \frac{0.4*0.05}{0.038} = 0.5263[/tex]
52.63% probability that this initial repair was made by the first technician
