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what is the probability of the document reaching its destination on time by exactly one of the services

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1 Answer

Hello Ibukunoluwa,
 
Let's call the events that the document reaches the destination on time by the first service, A, by the second service, B, and by the third service, C.  Then, the probability that you are interested in computing is the probability that only A or only B or only C occurs.  First, denote by
 
P(A ∪ B ∪ C),
 
the probability that the document reaches its destination on time by any of the services, and recall that
 
 
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C) + P(A ∩ B ∩ C),
 
by the so-called Inclusion-Exclusion Principle.  Now, it seems a reasonable assumption to make that the three delivery services operate independently of one another; in this case, we can simplify the above expression by making use of the definition of independence.  We may replace the probabilities of the intersections by the product of the probabilities of the events making up the intersections, to obtain:
 
 
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P(A)P(B) - P(A)P(C) - P(B)P(C) + P(A)P(B)P(C).
 
This is then an expression we can use, if we know the individual probabilities!  Since we are interested in the probability that the document arrives by exactly one of the services, then we want to compute the probability
 
 
 
P(A ∪ B ∪ C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C) + 2·P(A ∩ B ∩ C),
 
which is apparent from drawing a Venn diagram, for instance, or invoking the Inclusion-Exclusion Principle again.  Invoking independence again, and using the above equations, we find the desired probability is
 
P(A) + P(B) + P(C) - 2·P(A)P(B) - 2·P(A)P(C) - 2·P(B)P(C) + 3·P(A)P(B)P(C),
 
an expression which can be whittled down into even nicer forms if desired.
 
Regards,
Hassan H.

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I thought I might clean up the expression above and explain another way to look at it, since I have a bit of unexpected time to spare today.  If we let a = P(A), b = P(B), and c = P(C), then we can rewrite the probability as
 
a + b + c - 2ab -2ac -2bc + 3abc 
 
which can be further rewritten as
 
a(1 - b - c + bc) + b(1 - a - c + ac) + c(1 - a - b + ab) = a(1-b)(1-c) + b(1-a)(1-c) + c(1-a)(1-b).
 
Now, denote the complementary probabilities by: (1-a) = ac, (1-b) = bc, and (1-c) = cc.  Then, the probability takes the pleasant form
 
abccc + acbcc + acbcc.
 
Finally, think about how to derive the latter expression without going through the Inclusion-Exclusion process, i.e. derive it from principles of probability.  [Hint: The probability that only A occurs is equal to the product of the probability that A occurs, that B does not occur, and that C does not occur.]
 
Regards,
Hassan H.

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