Additive rules
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}
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
- 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:
return to index
Questions?