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When using the addition rule in probability, why must we subtract the "intersection" to find the "union" with the Addition Rule?

There is a subtle point to be made here, which comes down to double counting. 

The addition rule:

Pr(A union B) = Pr(A)+Pr(B)-Pr(A intersection B)

allows us to work out the probability that either event A or event B happens: i.e., it tells us for any two events A and B, what is the probability that one of them occurs. 

However, this probability is affected by the relationship between the two events themselves. In particular, it matters whether the events are mutually exclusive or not.

Consider the following example:

A class of 20 students contains 12 boys and 8 pupils with blonde hair. What is the probability that a student is either a boy or has blonde hair?

Let event A be that a student is chosen with blonde hair and let event B be that a boy is chosen. Thus we are trying to find Pr(A union B) - what is the probability of event A or B occuring? The natural response is to think it is simply Pr(A) + Pr(B), in this case 8/20+12/20 =20/20=1

However, this is only true if there are no boys with blonde hair. Suppose instead that there are 2 boys with blonde hair in the class. Now the probability of choosing a student that is either a boy or blonde has fallen, since of the 8 remaining girls in the class, 2 do not have blonde hair. So we must calculate:

Pr(A union B) = Pr(A)+Pr(B)-Pr(A intersection B)

Here, Pr(A intersection B) is the probability that a student is a blonde boy, which is 2/20. Therefore, our new probability is:

Pr(A union B)=8/20+12/20-2/20=18/20

If we did not subtract the term at the end, we would be double counting the blonde haired boys firstly as boys and then as students with blonde hair, fogetting that they are one and the same individual in 2 cases. 

To make this point another way, consider a Venn diagram. With two mutually exclusive events, two circles in the Venn diagram do not overlap. With non-mutually excusive events, the circles do overlap. The overlap is the intersection, calculated here as 2/20. If we do not subtract this intersection from Pr(A) +Pr(B), we double count it, giving us the wrong probability of both events happening.

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