P&S L2 C2 S4

When we add probabilities of two events which are not mutually exclusive, then we double count the overlap -- it is added in as part of A, and also as part of B. There are many ways to correct for this to get the right formula, We just subtract the probability of the intersection of the two sets from the sum:

EXAMPLE. Consider a situation where the males also start to get ears pierced, so the two sets of the previous slide are not mutually exclusive. Suppose we have 60 males, 40 females, 25 females with pierced ears, and 10 males with pierced ears. Let A be the set of males and B the set of people with pierced ears. P(A)=60% and P(B)=35%. However P(AUB) is NOT the sum of the probabilities, which is 60+35=95%. It is not the sum because when we add the two sets, we DOUBLE COUNT the males with pierced ears. If we subtract this double count, we can get the right result:

P(AUB)=60% +35% - 10% =85%

The second formula counts males WITHOUT pierced ears = 50%, Females with pierced ears=25% and The intersection of pierced ears with males, which is 10%. So we have

P(AUB)=50%+25%+10%=85% -- the same result

Another way to calculate the same sum is to put it into THREE pieces. One piece is A-B which the set A after removing elements of B. The second is B-A which is B with elements of A removed from it. Finally the third piece is the intersection, which is all members of both sets. These three sets are mutually exclusive and so we can add their probabilities. This gives the following formula