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) = 0.14 P(A2) = 0.09 P(A3) = 0.09 P(A1 ∪ A2) = 0.16 P(A1 ∪ A3) = 0.18 P(A2 ∪ A3) = 0.14 P(A1 ∩ A2 ∩ A3) = 0.02 (a) What is the probability that the system does not have a type 1 defect? (b) What is the probability that the system has both type 1 and type 2 defects? (c) What is the probability that the system has both type 1 and type 2 defects but not a type 3 defect? (d) What is the probability that the system has at most two of these defects?

Answer 1

a.

`P({A_1}^c)=1-P(A_1)=\boxed{0.86}`

b.

`P(A_1\cap A_2)=P(A_1)+P(A_2)-P(A_1\cup A_2)=\boxed{0.07}`

c. By the law of total probability,

`P({A_3}^c)=P((A_1\cap A_2)\cap{A_3}^c)+P((A_1\cap A_2)^c\cap{A_3}^c)`

According to the inclusion/exclusion principle,

`P((A_1\cap A_2)^c\cap{A_3}^c)=P((A_1\cap A_2)^c)+P({A_3}^c)-P((A_1\cap A_2)^c\cup{A_3}^c)`

but by DeMorgan's law,

`(A_1\cap A_2)^c\cup{A_3}^c={A_1}^c\cup{A_2}^c\cup{A_3}^c=(A_1\cap A_2\cap A_3)^c`

So

`P((A_1\cap A_2)^c\cap{A_3}^c)=(1-P(A_1\cap A_2))+(1-P(A_3))-(1-P(A_1\cap A_2\cap A_3))`

and from there we find

`P(A_1\cap A_2\cap{A_3}^c)=P((A_1\cap A_2\cap A_3)^c)-P((A_1\cap A_2)^c)=\boxed{0.05}`

d. The event of having at most two of these defects is complementary to the event that all three defects occur simultaneously:

`P((A_1\cap A_2\cap A_3)^c)=1-P(A_1\cap A_2\cap A_3)=\boxed{0.98}`