Question 1179809
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The health department of a city routinely conducts two independent inspections 
of each restaurant, with the restaurant passing only if both inspectors pass it.
Inspector A is very experienced and hence passes only 2 percent of restaurants
that actually do have health code violations. Inspector B is less experienced
and pass 7 percent of restaurants with health code violations what is the
probability that

(a)  Inspector A passes a restaurant, give that inspector B has found a violation?

(b)  Inspector B passes a restaurant with a violation given that inspector A passes it? 
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            This problem is on conditional probability.



<pre>
(a)  This question is to find the conditional probability

                                             P(A passes AND B found a violation)
        P(A passes | B found a violation) = --------------------------------------
                                                   P(B found a violation)


     These events, (A passes) and (B found a violation) are independent, so the probability
     of their intersection is the product of probabilities

       P(A passes AND B found a violation) = P(A passes) * P(B found a violation) = 0.02*(1-0.07) = 

           = 0.02*0.93.


     So, the conditional probability is  P(A passes | B found a violation)) = {{{(0.02*0.93)/0.93}}} = 0.02.

     This is the <U>ANSWER</U>  to question (a).




(b)  For question (b), the logic/(the reasoning) is similar, and it leads to the solution

         
                                             P(B passes AND A passes)      0.07*0.02
        P(B passes | A found a violation) = -------------------------- = --------------- = 0.07.
                                                  P(A passes)                 0.02


     This is the <U>ANSWER</U>  to question (b).


The answers are consistent with the intuitive feeling / understanding of independent events.
</pre>

Solved.