1. I know that customers who live in a geographical area (A) purchase an item 75% of the time. 2. I know that customers who have an income (I) purchase an item 60% of the time. 3. I have a customer (C) with the characteristics of (A) and (I). 4. I\'m not aware of any significant correlation or influencing factors between populations (A) and (I), and the data sample for that overlapping is not large enough-- Lets say Customer (C) is the only member of the population with both (A) and (I). How would I determing the liklihood that Customer (C) will purchase an Item? Would averaging the probabilities be appropriate, for (0.6+0.75)/2 = 0.675%? Solution First of all, the problem is not well posed, since the formulation is contradicting. If everyone in A purchases 75% of the time (or with probability 3/4) and everyone with income I purchases 60% of the time (or with probability 3/5), then there can be no overlap, since no single person can buy with two different probabilities (or 75% of the time *and* 60% of the time). Instead, you could consider the following scenario: Let\'s say a new product is on the market. Among the #A people living in A, 75% decide to buy it. Among the #I people with income I, 60% decide to buy it. What fraction of people living in A with income I decide to buy? Let this fraction be P. Let S be the intersection of A and I, i.e., the set of people living in A who have income I. This question assumes that #S>0. While this is a well-posed problem, it is not possible to say much about it, since the answer depends on the relative size of the sets. We can only give ranges: Let A\\I be the set of people in A whose income is not I. Let I\\A be the set of people with income I who don\'t live in A. If the intersection #S is small relative to both #A and #I the answer could be anything from 0% to 100%. Since 75% of the folks in A buy: If #S>#A/4, then P>0. (Even if everyone in A\\I bought, some folks in S need to buy, too.) If #S>#A 3/4, P<1. (Even if everyone in A\\I didn\'t buy, some folks in S didn\'t buy, either.) Since 60% of the folks in I buy: If #S>#I 2/5, P>0. (Even if everyone in I\\A bought, some folks in I need to buy, too.) If #S>#I 3/5, P<1. (Even if everyone in I\\A didn\'t buy, some folks in S didn\'t buy, either.) Now, let\'s generalize this. Let x=#S/#A and y=#S/#I. We can express the feasible region (admissible values for P) as follows: For A: We have the following constraints: P>=0, P<=1, P>=x-1/4, P<=7/4-x. For I: We the constraints: P>=0, P<=1, P>=y-2/5, P<=8/5-y. When you draw theses bounds on P as a function of x and y, you get two pentagonal areas. P has to lie in both of them. If x and y are given, this allows you to find the admissible values of P. You see that for x=1, only P=3/4 is possible. For y=1, only P=3/5 is possible. So x=y=1 is not possible, this would mean that A=I=S, which obviously would violate the assumption. You could ask the question what is the largest (common) value of x and y such that a so.