To illustrate the additive rules, we shall consider the probability space with probabilities of outcomes as in the table below:
```      outcome:    |   r   |   s   |   t   |   u   |
----------------------------------------------
probability:|  .1   |  .4   |  .2   |  .3   |
```
Let A={r, s}; B={s, t}; C={u}

The probability of an event is the sum of the probabilities in the outcomes in the event:
P(A)=.1+.4=.5
P(B)=.4+.2=.6
P(C)=.3
P(AUB)=.1+.4+.2=.7, since AUB={r, s, t}
P(AB)=.4, since AB={s}
P(B')=.4, since B'={r, u}

## Additive rule for disjoint (mutually exclusive) events

Two events X and Y are called disjoint or mutually exclusive if XY=Ø ({}, the empty or null set), i.e., X and Y do not share any outcome.
P(XUY)=P(X)+P(Y) if X and Y are disjoint events. (The proof of this is just the associative rule of addition.)
P(AUC)=.8=.5+.3=P(A)+P(C) since AC= Ø
P(AUB)=.7 does not equal .5+.6=P(A)+P(B) since AB={s} which is not Ø

The reason P(AUB) is not equal to P(A)+P(B) is that the outcomes in the intersection of A and B (i.e., {s}) are counted twice when you add the probabilities. The general additive rule is

P(XUY) = P(X) + P(Y) - P(XY),
which in the case of A and B gives
.7 = P(AUB) = P(A) + P(B) - P(AB) = .5 + .6 - .4

## Rules for complements

• P(A') + P(A) = 1
• P(AA') = 0
• P(A) = 1- P(A')
Note that the definition of complementary events is that AUA'=S (the entire sample space) *and* AA'= Ø (the null set).

Competencies: If P(A)=.6, P(B)=.5, and P(AB)=.2; P(A')=?, P(AUB)=?
If P(A)=.5, P(AB)=.3, and P(AUB)=.8, P(B)=?
If P(A)=.3 an A and B are complementary, P(B)=?
If P(A)=.3, P(B)=.5, and P(AUB)=.8, are A and B mutually exclusive (disjoint)?

Challenge: