?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 haveP(X) = 0.7P(Y') = 0.5from whichP(X') = 0.3P(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, asP(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 writtenP(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)] = 0.6orP[(X?Y)] = 1 - 0.6 = 0.4.From the equation aboveP(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 assumptionP(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) = 0.85which is slightly di