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14 July, 01:51

Suppose that 95% of the fasteners pass the initial inspection. Of those that fail the initial inspection, 20% are defective. Of the fasteners sent to the recrimping operation, 40% cannot be corrected ad are scrapped; the are corrected by the recrimping and then pass inspection.

A. What proportion of the fasteners that fail the initial inspection pass the second inspection (after the recrimping operation) ?

B. What proportion of fasteners pass inspection?

C. Given that a fastener passes inspection, what is the probability that it passed the initial inspection and did not have to go through the recrimping operation?

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  1. 14 July, 02:03
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    Answer

    Step-by-step explanation:

    Let x be the fasteners

    95% of x passed, and 5% of x failed

    20% of the 5% which failed are defective, remaining the 80% of the failed 5%.

    Now 40% of this 80% of the 5% which failed are scrapped, meaning only the remaining 60% passed second inspection.

    Thus

    a). The proportion;

    60% of 80% of 5% of x

    mathematically

    = 60/100 * 80/100 * 5/100

    = 0.024 or 2.4%

    b) initially 95% passed and later on 2.4% pass again. Therefore total that passed = 95%+2.4%=97.4%

    c) probability that it passed on first inspection is 95/100

    Percentage that went through recrimping = 80% of 5% = 80/100 * 5/100 = 1/25 or 0.04 or 4%

    Thus probability of passing and not going through recrimping

    = 1 - (1/25) = 0.96 or 96%
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