A certain system can experience three different types of defects. Let Ai (i = 1,

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) = 0.10 P(A2) = 0.08 P(A3) = 0.05 P(A1 A2) = 0.12 P(A1 A3) = 0.12 P(A2 A3) = 0.11 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.)

b) Given that the system has a type 1 defect, what is the probability that it has all three types of defects? (Round your answer to four decimal places.)

c) Given that the system has at least one type of defect, what is the probability that it has exactly one type of defect? (Round your answer to four decimal places.)

d) Given that the system has both of the first two types of defects, what is the probability that it does not have the third type of defect? (Round your answer to four decimal places.)

Explanation / Answer

A1, A2, A3 are the three defects.

Make 3 circles intersecting each other (like set theory) with a1 intersecting a2 intersecting a3 in the middle.

a)

p (a2/ a1) = P (a2 intersecn a1)/ p(a1)

a1 + a2 - a1 union a2 = a1 intersecn a2

0.1+0.08-0.12 = 0.06

p (a2/ a1) = 0.06/0.1 =0.6

b)

p(a1 intersecn a2 intersecn a3/ a1) = p(a1 intersecn a2 intersecn a3 intersecn a1) / P(a1)

= p(a1 intersecn a2 intersecn a3 ) / P(a1)

=0.01/0.1 = 0.1000

c)

a1 only or a2 only or a3 only

= A1+A2-A1 intersecn A2 - A1 intersecn A3 + a1 intersecn a2 intersecn a3

=0.1+0.08-0.06-0.03+0.01=0.1

d)

P(not a3/ a1 intersecn a2) = a1 intersecn a2 - a1 intersecn a2 intersecn a3 = 0.06-0.01 = 0.05

A1 0.1 A2 0.08 A3 0.05 A1 UNION A2 0.12 A2 UNION A3 0.11 A3 UNION A1 0.12 A1 INTERSECN A2 INTERSECN A3 0.01 A1 INTERSECN A2 0.06 A1 INTERSECN A3 0.03 A2 INTERSECN A3 0.02

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