Question

A certain system can experience three different types of defects. Let Ai (i = 1,2,3) denote the event that the system has a defect of type i. Suppose that the following probabilities are true. P(A1) = 0.12 P(A2) = 0.08 P(A3) = 0.06 P(A1 ∪ A2) = 0.14 P(A1 ∪ A3) = 0.15 P(A2 ∪ A3) = 0.12 P(A1 ∩ A2 ∩ A3) = 0.01 (a) Given that the system has a type 1 defect, what is the probability that it has a type 2 defect? (Round your answer to four decimal places.)

Answer 1

By definition of conditional probability,

`P(A_2\mid A_1)=(P(A_1\cap A_2))/(P(A_1))`

The inclusion-exclusion principle tells us

`P(A_1\cap A_2)=P(A_1)+P(A_2)-P(A_1\cup A_2)=0.12+0.08-0.14=0.06`

Then

`P(A_2\mid A_1)=(0.06)/(0.12)=0.5000`