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

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) = 0.05P(A1 ? A2) = 0.14 P(A1 ? A3) = 0.14P(A2 ? A3) = 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) 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.)

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Ella 2 months 2021-10-18T07:40:33+00:00 1 Answer 0 views 0

Answers ( )

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    2021-10-18T07:42:22+00:00

    Answer:

    (a) 0.5000

    (b) 0.0833

    (c) 0.2500

    ((d) 0.9500

    Step-by-step explanation:

    Given that the following probabilities;

    P(A1) = 0.12; P(A2) = 0.08,

    P(A3) = 0.05; P(A1 ∪ A2) = 0.14, P(A1 ∪ A3) = 0.14,

    P(A2 ∪ A3) = 0.11,

    P(A1 ∩ A2 ∩ A3) = 0.01

    (a) we want to find P(A2/A1).

    P(A2/A1) = P(A1nA2)/P(A1)……….a

    So we need to find P(A1nA2)

    P(A1 ∪ A2)= P(A1) + P(A2) – P(A1nA2)

    P(A1nA2) =P(A1) + P(A2) – P(A1 ∪ A2)

    P(A1nA2) = 0.12 + 0.08 – 0.14

    P(A1nA2) = 0.20- 0.14

    P(A1nA2 )= 0.060

    P(A2/A1) = 0.06/0.12

    = 0.5000. (Ans)

    (b) The required probability is: P(A2∩A3/A1) =[P(A2∩A3)∩P(A1)] /P(A1)………b

    But [P(A2nA3)nP(A1)] = P(A1∩A2∩ A3)

    P(A2∩A3/A1) = 0.01/0.12

    P(A2∩A3/A1) = 0.0833 (Ans)

    (c) the required probability is given by Pr =

    (Pr. of A1 defect) + (Pr. of A2 defect) + (Pr. of A3 defect)……..c

    Pr = 0.12 + 0.08 + 0.05

    Pr = 0.2500 (And)

    (d) The required probability is given by:

    Pr[(A3)’ /(A1∩A2)] = Pr[(A3)’nPr(A1∩A2)] /Pr(A1∩A2)]

    Pr(A3)’ = 1 – Pr(A3) = 1 -0.05

    Pr(A3)’/(A1∩A2)= (0.95×0.06)/(0.06)

    Pr(A3)’/(A1∩A2 )= 0.95(Ans)……d

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