outcome: | r | s | t | u | ---------------------------------------------- probability:| .1 | .4 | .2 | .3 |Let A={r, s}; B={s, t}; C={u}

- Additive rule for outcomes
- Additve rule for disjoint (mutually exclusive) events
- General additive rule
- Rules for complements

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}

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

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

- P(A') + P(A) = 1
- P(AA') = 0
- P(A) = 1- P(A')

**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:**