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A chartered accountant applies for a job in two firms X and Y. He estimates that the probability of his being selected in firm X is 0.7 and beig rejected in Y is 0.5 and the probability that atleast one of his applications rejected is 0.6. What is the probability that he will be selected in one of the firms?

Naveen,

Let the event X = selected at company X and the event Y = selected at company Y. This means that X' = rejected at X and Y' = rejected at Y. Here the prime notation means the "complement of". From the information given, we have

P(X) = 0.7

P(Y') = 0.5

from which

P(X') = 0.3

P(Y) = 0.5.

using P(X) = 1 - P(X'), etc.

As stated, we want the probability of the event X∪Y = union of events X and Y = either X or Y, or both, happens (one or the other or both of the applications being selected). In other words, we want P(X∪Y), which can be written, using the axioms of probability theory, as

P(X∪Y) = P(X) + P(Y) - P(X∩Y) where the ∩ symbol means the "intersection of" or, in other words, both X and Y happen. So now we need to calculate P(X∩Y).

From the information given, we know that the probability of either X' or Y', or both, happening is 0.6, which can be written

P(X'∪Y') = 0.6.

From the axioms of set theory we have (X'∪Y') = (X∩Y)' = complement of the intersection of X and Y (complement of both X and Y happening). We also have from probability theory that P(event) = 1 - P(event'), so that

P[(X∩Y)'] = P[(X'∪Y')] = 1 - P[(X∩Y)] = 0.6

or

P[(X∩Y)] = 1 - 0.6 = 0.4.

From the equation above

P(X∪Y) = P(X) + P(Y) - P(X∩Y) = 0.7 + 0.5 - 0.4 = 0.8, which is your answer.

Note that this means that the events X and Y are not independent since, if we make that assumption

P(X∪Y) = P(X) + P(Y) - P(X∩Y) = P(X) + P(Y) - P(X)･P(Y) = 0.7 + 0.5 - (0.7)(0.5) = 0.85

which is slightly different.

Hope this makes sense.

Randy

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