SOLUTION: Suppose that in the maintenance of a large medical records file for insurance purposes the probability of an error in processing is 0.0010, the probability of an error in filing is

In this problem, you are given three subsets


    C = errors in proCessing          ( 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 AND retrieving   ( P(CF) = 0.0002 )

    CR = errors in processing AND retrieving   ( P(CR) = 0.0003 )

    FR = errors in filing AND retrieving       ( P(FR) = 0.0002 )

    CFR = errors in processing AND filing AND 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 ANSWER 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.