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) = 0.11

P(A2) = 0.08

P(A3) = 0.05

P(A1 A2) = 0.13

P(A1 A3) = 0.13

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

as we know that P(AnB) =P(A)+P(B)-P(AUB)

hence P(A1 nA2) =0.06

P(A1nA3) =0.03

P(A2nA3) =0.02

a) probability that it has a type 2 defect ; given system has a type 1 defect =P(A2|A1)=P(A1nA2)/P(A1)

=0.06/0.11 =0.5455

b)P(A1nA2nA3|A1) =P(A1nA2nA3)/P(A1) =0.01/0.11 =0.0909

c)probability of at least one type of defect

=P(A1uA2UA3)=P(A1)+P(A2)+P(A3)-P(A1nA2)-P(A2nA3)-P(A1nA3)+P(A1nA2nA3)=0.14

and probability of exactly one type of defect

=P(A1uA2UA3)=P(A1)+P(A2)+P(A3)-2*P(A1nA2)-2*P(A2nA3)-2*P(A1nA3)+3*P(A1nA2nA3) =0.05

  probability that it has exactly one type of defect given at least one type of defect

=0.05/0.14=0.3571

d)probability =(P(A1nA2)-P(A1nA2nA3))/P(A1nA2) =(0.06-0.01)/0.06=0.8333

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.