PROBABILITY
This is particularly useful for problems in which evaluating the probability
of the complement is easier than evaluating the probability of the event
itself.Calculate the probability of drawing an ace or a spade from a pack of cards.LetAbe the event that an ace is drawn andBthe event that a spade is drawn. It
immediately follows that Pr(A)= 524 = 131 and Pr(B)=^1352 =^14. The intersection ofAand
Bconsists of only the ace of spades and so Pr(A∩B)= 521. Thus, from (30.9)
Pr(A∪B)= 131 +^14 − 521 = 134.In this case it is just as simple to recognise that there are 16 cards in the pack that satisfy
the required condition (13 spades plus three other aces) and so the probability is^1652 .
The above theorems can easily be extended to a greater number of events. Forexample, ifA 1 ,A 2 ,...,Anare mutually exclusive events then (30.10) becomes
Pr(A 1 ∪A 2 ∪···∪An)=Pr(A 1 )+Pr(A 2 )+···+Pr(An). (30.12)Furthermore, ifA 1 ,A 2 ,...,An(whether mutually exclusive or not)exhaustS,i.e.
are such thatA 1 ∪A 2 ∪···∪An=S,then
Pr(A 1 ∪A 2 ∪···∪An)=Pr(S)=1. (30.13)A biased six-sided die has probabilities^12 p,p,p,p,p, 2 pof showing1, 2, 3, 4, 5, 6
respectively. Calculatep.Given that the individual events are mutually exclusive, (30.12) can be applied to give
Pr(1∪ 2 ∪ 3 ∪ 4 ∪ 5 ∪6) =^12 p+p+p+p+p+2p=^132 p.The union of all possible outcomes on the LHS of this equation is clearly the sample
space,S,andso
Pr(S)=^132 p.Now using (30.7),
13
2 p=Pr(S)=1 ⇒ p=
2
13 .When the possible outcomes of a trial correspond to more than two events,and those events arenotmutually exclusive, the calculation of the probability of
the union of a number of events is more complicated, and the generalisation of
the addition law (30.9) requires further work. Let us begin by considering the
union of three eventsA 1 ,A 2 andA 3 , which need not be mutually exclusive. We
first define the eventB=A 2 ∪A 3 and, using the addition law (30.9), we obtain
Pr(A 1 ∪A 2 ∪A 3 )=Pr(A 1 ∪B)=Pr(A 1 )+Pr(B)−Pr(A 1 ∩B).
(30.14)