Comments on the General Addition rule

Actually it works for both, when events are disjoint P(AB)= 0. (in other words since they are disjoint, they don’t overlap so the probability of both A and B happening is zero). And this is the addition rule that works for non-disjoint events, because you have to subtract the overlap, otherwise you are counting that part twice. For example, lets say there are 50 students taking AP stats at East and West. Of those 50 lets define event A as being on an academic team. And event B as being on an athletic team. These events will likely overlap since some people just get involved in everything. For this example lets say there are 20 students out of the 50 on an academic team, 10 on an athletic team, and 4 on both. To determine the probability of being in on an athletic team or an academic team, we need to know the total students on athletic teams or academic teams. So 20 + 10, this could also be written as (16 + 4) + (6 + 4), the 16 are students only on an academic team, the 4 are on both, the 6 are only on an an athletic team the 4 are on both. But the problem is the 4 are the same 4 in each parenthesis. We should not count them twice. So we would do (16+4)+(6+4)-4 or 20+10-4 to get 26 students in athletics or academics, the probability would be 2650= .52 This was done with counts but the same could have been done with percents (.4) + (.2) - .8 = .52.