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,

$P(A_2\mid A_1)=\dfrac{P(A_1\cap A_2)}{P(A_1)}$

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

$P(A_2\mid A_1)=\dfrac{0.06}{0.12}=0.5000$