7. 4 points DevoreStat9 2.E 048 My Notes Ask Your Teacher A certain system can e

Question

7. 4 points DevoreStat9 2.E 048 My Notes Ask Your Teacher A certain system can experience three different types of defects. Let A, (1,2,3) denote the event that the system has a defect of type i. Suppose that the following probabilities are true. P(A;)= 0.12 P(42)=0.08 P(An)=0.05 RAI UA2) = 0.14 P(A1 UA3) = 0.14 PLA2 U Az) = 0.11 P(A1 n A2 n 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) Glven that the system has at least one type of defect, what is the probablity that it has exactly one type of defect? (Round your answer to four decimal places.) (d) Glven that the system has both of the first two types of defects, what Is the probablity that it does not have the third type of defect? (Round your answer to four declmal places.)

Explanation / Answer

(a) Here writing question mathematically,

Pr(A2U A1) = Pr(A1) + Pr(A2) -  Pr(A1 A2)

0.14 = 0.12 + 0.08 -  Pr(A1 A2)

Pr(A1 A2) = 0.06

Pr(A2 l A1) = Pr(A2 A1)/Pr(A1) = 0.06/0.12 = 0.5000

(b) Pr(A1 A2 A3 l A1) =Pr(A1 A2 A3 )/ Pr(A1) = 0.01/0.12 = 1/12 = 0.0833

(c) Given that system has at least one type of defect that means union of all these defects

Pr(A1 U A2 U A3) = Pr(A1U A2) + Pr(A2U A3) + Pr(A1U A3) - (Pr(A1) + Pr(A2) + Pr(A3) ) + Pr(A1 A2 A3)

= 0.14 + 0.14 + 0.11 - 0.12 - 0.08 -0.05 + 0.01 = 0.15

Pr(Exactly one type of defect) = 2* [Pr(A1U A2) + Pr(A2 U A3) + Pr(A1 U A3) ] - 3 [ (Pr(A1) + Pr(A2) + Pr(A3) ] + 3 * Pr(A1 A2 A3) = 2 * 0.39 - 3 * 0.25 + 3 * 0.01 = 0.0600

Pr(Exactly one type of defect if there is an defect) = 0.06/0.15 = 0.4000

(d) System has both type of defects

Pr(Both type of defects) = Pr(A1 A2) = 0.06

Pr(It doesn't have third type of defect = 0.06 - Pr(A1 A2 A3) = 0.06 - 0.01 = 0.05

Pr(Doesn't have type III defect l have first two defect) = 0.05/0.06 = 5/6 = 0.8333

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