Components arriving at a distributor are checked for defects by two different inspectors (each component is checked by both inspectors). The

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Components arriving at a distributor are checked for defects by two different inspectors (each component is checked by both inspectors). The first inspector detects 83% of all defectives that are present, and the second inspector does likewise. At least one inspector does not detect a defect on 34% of all defective components. What is the probability that the following occur

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Kim Cúc 4 years 2021-08-07T05:37:04+00:00 1 Answers 97 views 0

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  1. King
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    2021-08-07T05:38:35+00:00
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    Complete question is;

    Components arriving at a distributor are checked for defects by two different inspectors (each component is checked by both inspectors). The first inspector detects 83% of all defectives that are present, and the second inspector does likewise. At least one inspector does not detect a defect on 34% of all defective components. What is the probability that the following occurs?

    (a) A defective component will be detected only by the first inspector?

    b) A defective component will be detected by exactly one of the two inspectors?

    (c) All three defective components in a batch escape detection by both inspectors (assuming inspections of different components are independent of one another)?

    Answer:

    A) 0.17

    B) 0.34

    C) 0

    Step-by-step explanation:

    a) We are told that the first inspector(A) detects 83% of all defectives that are present, and the second inspector(B) also does the same.

    This means that;

    P(A) = P(B) = 83% = 0.83

    We are also told that at least one inspector does not detect a defect on 34% of all defective components.

    Thus;

    P(A’ ⋃ B’) = 0.34

    Also, we now that;

    P(A ⋂ B) = 1 – P(A’ ⋃ B’)

    P(A ⋂ B) = 1 – 0.34

    P(A ⋂ B) = 0.66

    Probability that A defective component will be detected only by the first inspector is;

    P(A ⋂ B’) = P(A) – P(A ⋂ B)

    P(A ⋂ B’) = 0.83 – 0.66

    P(A ⋂ B’) = 0.17

    B) probability that a defective component will be detected by exactly one of the two inspectors is given as;

    P(A ⋂ B’) + P(A’ ⋂ B) = P(A) + P(B) – 2P(A ⋂ B)

    P(A) + P(B) – 2P(A ⋂ B) ; 0.83 + 0.83 – 2(0.66) = 0.34

    C) Probability that All three defective components in a batch escape detection by both inspectors is written as;

    P(A’ ⋃ B’) – (P(A ⋂ B’) + P(A’ ⋂ B))

    Plugging in the relevant values, we have;

    0.34 – 0.34 = 0

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