Mathematical Methods for Physics and Engineering : A Comprehensive Guide

PROBABILITY

SinceA∩X=Xwe must haveX⊂A. Now, using the second equality in (30.4) in a
similar way, we find

A∪X=A∪[A∩(A∪B)]

=(A∪A)∩[A∪(A∪B)]

=A∩(A∪B)=X,

from which we deduce thatA⊂X.Thus,sinceX⊂AandA⊂X, we must conclude that
X=A.
(ii) Since we do not know how to deal with compound expressions containing a minus
sign, we begin by writingA−B=A∩B ̄as mentioned above. Then, using the distributivity
law, we obtain

(A−B)∪(A∩B)=(A∩B ̄)∪(A∩B)

=A∩(B ̄∪B)

=A∩S=A.

In fact, this result, like the first one, can be proved trivially by drawing a Venn diagram.

Further useful results may be derived from Venn diagrams. In particular, it is

simple to show that the following rules hold:

(i) ifA⊂BthenA ̄⊃B ̄;

(ii)A∪B=A ̄∩B ̄;

(iii)A∩B=A ̄∪B ̄.

Statements (ii) and (iii) are known jointly asde Morgan’s lawsand are sometimes

useful in simplifying logical expressions.

There exist two eventsAandBsuch that

(X∪A)∪(X∪A ̄)=B.

Find an expression for the eventXin terms ofAandB.

We begin by taking the complement of both sides of the above expression: applying de
Morgan’s laws we obtain

B ̄=(X∪A)∩(X∪A ̄).

We may then use the algebraic laws obeyed by∩and∪to yield

B ̄=X∪(A∩A ̄)=X∪∅=X.

Thus, we find thatX=B ̄.

30.2 Probability

In the previous section we discussed Venn diagrams, which are graphical repre-

sentations of the possible outcomes of experiments. We did not, however, give

any indication of how likely each outcome or event might be when any particular

experiment is performed. Most experiments show some regularity. By this we

mean that the relative frequency of an event is approximately the same on each

occasion that a set of trials is performed. For example, if we throw a dieN