SOLUTION: 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 ve

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Question 1179809: 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?

Answer by ikleyn(52786) About Me  (Show Source):
You can put this solution on YOUR website!
.
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?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


            This problem is on conditional probability.


(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)) = %280.02%2A0.93%29%2F0.93 = 0.02.

     This is the ANSWER  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 ANSWER  to question (b).


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

Solved.