Items are inspected for flaws by two quality inspectors. If a flaw is present, it will be detected by the first inspector with probability 0

Question

Items are inspected for flaws by two quality inspectors. If a flaw is present, it will be detected by the first inspector with probability 0.93, and by the second inspector with probability 0.7. Assume the inspectors function independently. If an item has a flaw, what is the probability that it will be found by at least one of the two inspectors

a. If an item has a flaw, what is the probability that it will be found by both inspectors?
b. If an item has a flaw, what is the probability that it will be found by at least one of the two inspectors?
c. Assume that the second inspector examines only those items that have been passed by the first inspector. If an item has a flaw, what is the probability that the second inspector will find it?

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Athena 3 months 2022-02-10T06:31:45+00:00 1 Answer 0 views 0

Answers ( )

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    2022-02-10T06:33:30+00:00

    Answer:

    (a) 0.651

    (b) 0.979

    (c) 0.049

    Step-by-step explanation:

    Let A = The flaw is detected by the first inspector and B = The flaw is detected by the second inspector.

    Given:

    P (A) = 0.93 and P(B) = 0.70

    Also A and B are independent, i.e P (A ∩ B) = P (A) × P (B)

    (a)

    Compute the probability that the flaw will be found by both inspectors as follows:

    [tex]P(Flaw\ will\ be\ found\ by\ both)=P(A)\times P(B)\\=0.93\times0.70\\=0.651[/tex]

    Thus, the probability that the flaw will be found by both inspectors is 0.651.

    (b)

    Compute the probability that the flaw will be detected by at least one of the two inspectors
    :

    [tex]P (At\ least\ one\ detected\ the\ flaw)=P(A)\timesP(B^{c})+P(A^{c})\timesP(B)+P(A\cap B)\\=[P(A)\times(1-P(B))]+[(1-P(A))\times P(B)]+[P(A)\times P(B)]\\=[0.93\times0.30]+[0.07\times0.70]+[0.93\times0.70]\\=0.979[/tex]

    Thus, the probability that the flaw will be found by at least one of the two inspectors is 0.979.

    (c)

    Compute the probability that the approved item of the first inspector will be rejected by the second as follows:

    P (Accepted by first but rejected by second) = P (Accepted by first) ×

                                                                                    P (Rejected by second)

                                                                              [tex]=[1-P(A)]\times P(B)\\=(1-0.93)\times 0.70\\=0.049[/tex]

    Thus, the  probability that the approved item of the first inspector will be rejected by the second is 0.049.

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