Disjunctive Probabilities
When events are not mutually exclusive, the rule for calculating disjunctive probabilities becomes more complicated. Suppose, for example, that ex- actly half the class is female and exactly half the class is over nineteen and the age distribution is the same for females and males. What is the probabil- ity that a randomly selected student will be either a female or over nineteen? If we simply add the probabilities (1/2 + 1/2 � 1), we would get the result that we are certain to pick someone who is either female or over nineteen. But that answer is wrong, because a quarter of the class is male and not over nineteen, and one of them might have been randomly selected. The correct answer is that the chances are 3/4 of randomly selecting someone who is ei- ther female or over nineteen. It is easy to see that in 75 percent of the cases, a randomly selected student will be either female or over nineteen. The table also shows what went wrong with our initial calculation. The top row shows that 50 percent of the students are female. The left column shows that 50 percent of the students are over nineteen. But we cannot simply add these figures to get the proba- bility of a randomly selected student being either female or over nineteen. Why not? Because that would double-count the females over nineteen. We would count them once in the top row and then again in the left column. To compensate for such double-counting, we need to subtract the students who are both female and over nineteen. The upper left figure shows that this is 25%. So the correct way to calculate the answer is 50% + 50% – 25% = 75%.
