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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.
Solved.
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As you see, the solution requires a long explanation;
but if you know the general formula (1), the solution is elementary.