Question 1132996
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In this problem, you are given three subsets


    C = errors in pro<U>C</U>essing          ( P(C) = 0.0010 )

    F = errors in filing              ( P(F) = 0.0009 )

    R = errors in retrieving          ( P(R) = 0.0012 )


and their intersections, subsets


    CF = errors in processing <U>AND</U> retrieving   ( P(CF) = 0.0002 )

    CR = errors in processing <U>AND</U> retrieving   ( P(CR) = 0.0003 )

    FR = errors in filing <U>AND</U> retrieving       ( P(FR) = 0.0002 )

    CFR = errors in processing <U>AND</U> filing <U>AND</U> retrieving  ( P(CFR) = 0.0001 ).



From the Probability theory, if you are given three sets of events C, F, R and their in-pair 
intersections  CF, CR, FR  and their in-triple  intersection CFR,  then

    P( C U F U R ) = P(C) + P(F) + P(R) - P(CF) - P(CR) - P(FR) + P(CFR).                        (1)



So, apply this formula and substitute there all given data. You will get

     the probability of making at least one of these errors = 

     = P( C U F U R ) = 0.0010 + 0.009 + 0.0012 - 0.0002 - 0.0003 - 0.0002 + 0.0001 = 0.0106.    (2)


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



The answer to question (b) is the COMPLEMENT to (2)

    the probability of making none of these errors = 1 - P( C U F U R ) = 1 - 0.0106 = 0.9894.
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Solved.


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As you see, &nbsp;the solution requires a long explanation;


but if you know the general formula &nbsp;(1), &nbsp;the solution is &nbsp;<U>elementary</U>.