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The question is:

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 found 724 that were judged defective, inspector B found 751 such joints, and 1159 of 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?

(b).What is the probability that the selected joint was judged to be defective by inspector B but not by inspector A?

Is the the number 724 strictly in $A$; how about 751, is that strictly in $B$, and it is not in $A \cap B$ or in $A \cup B$?

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Disregard comment. It was saved in error. –  lewellen Jan 29 '13 at 16:19
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@lewellen No, rather $|A \cup B| = 1159$. $| A\cap B| = |A| + |B| - | A \cup B | = 316$. –  Sasha Jan 29 '13 at 16:20
    
Okay, I was able to solve part (a), but (b) is giving me a little trouble. From what I read, the cardinality of the event (set) that contains the outcomes (elements) where only inspector B finds a joint defective should be $|B \cup A'| = 751$, the probability being $P(|B \cup A'|) = 751/10,000 = .0751$. The true answer is, .0435. What did I do wrong? –  Mack Jan 29 '13 at 16:47
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$A$: the set of joints deemed defective by inspector A.

$B$: the set of joints deemed defective by inspector B.

$U$: the set of all joints.

The problem says $|A| = 724$, $|B| = 751$, $|A\cup B| = 1159$, $|U| = 10000$.

(a) This asks for the number of joints that are $\notin (A\cup B)$.

(b) This asks for the number of joints that are in B but not in A, which I represent as the set $B-A$.

Then, (a) is easy, $\frac{10000-1159}{10000}$ (which I guess you solved as well).

For (b), your expression is wrong, $B-A$ is not $B \cup A'$, but is rather $B \cap A'$ (You may want to draw a Venn diagram to see that is the case). Since we know $|A\cap B| = 316$, we then get $\frac{751-316}{10000}$.

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