Question

Human visual inspection of solder joints on printed circuit boards can be very subjective. Part of the problem stems from the numerous types of solder defects (e.g., pad non-wetting, knee visibility, voids) and even the degree to which a joint possesses one or more of these defects. Consequently, even highly trained inspectors can disagree on the disposition of a particular joint. In one batch of 10,000 joints, inspector A found742that were judged defective, inspector B found745such joints, and1145of the joints were judged defective by at least one of the inspectors. Suppose that one of the 10,000 joints is randomly selected.
(a) What is the probability that the selected joint was judged to be defective by neither of the two inspectors? (Enter your answer to four decimal places.)


(b) What is the probability that the selected joint was judged to be defective by inspector B but not by inspector A? (Enter your answer to four decimal places.)

Answer 1

Answer:

a

The probability that the selected joint was judged to be defective by neither of the two inspectors is   $P(A' n B' ) = 0.8855$

b

The probability that the selected joint was judged to be defective by inspector B but not by inspector A  is  $P(A' n B) =0.0403$

Step-by-step explanation:

From the question we are told that

   The sample size is $n_s = 10000$

    The number of outcome for inspector A is  $n__{A}} = 742$

    The number of outcome for inspector B is  $n__{B}} = 745$

     The number of joints judged defective by both inspector is $n(A u B) = 1145$

The the probability that the selected joint was judged to be defective by neither of the two inspectors is mathematically represented as

      $P(A' n B' ) = 1 - P(A u B)$

Now

       $P(A\ u \ B) = \frac{n(Au B)}{n_s }$

substituting values

        $P(A\ u \ B) = \frac{1145}{ 10 000 }$

So  

      $P(A' n B' ) = 1 - \frac{1145}{10 000}$

     $P(A' n B' ) = 0.8855$

the probability that the selected joint was judged to be defective by inspector B but not by inspector A  is mathematically represented as

     $P(A' n B) = P(A \ u \ B) -P(A)$

Now

        $P(A) = \frac{n__{A}}{n_s}$

substituting values

       $P(A) = \frac{742}{10 000}$

So

     $P(A' n B) = \frac{1145}{10 000} - \frac{742}{10 000}$

    $P(A' n B) =0.0403$