Mathematical Methods for Physics and Engineering : A Comprehensive Guide

30.2 PROBABILITY

ace from a pack of cards from which one has already been removed, given that

eventA, the card already removed was itself an ace, has occurred.

We denote this probability by Pr(B|A) and may obtain a formula for it by

considering the total probability Pr(A∩B)=Pr(B∩A) that bothAandBwill

occur. This may be written in two ways, i.e.

Pr(A∩B)=Pr(A)Pr(B|A)

=Pr(B)Pr(A|B).

From this we obtain

Pr(A|B)=

Pr(A∩B)

Pr(B)

(30.19)

and

Pr(B|A)=

Pr(B∩A)

Pr(A)

. (30.20)

In terms of Venn diagrams, we may think of Pr(B|A) as the probability ofBin

the reduced sample space defined byA. Thus, if two eventsAandBare mutually

exclusive then

Pr(A|B)=0=Pr(B|A). (30.21)

When an experiment consists of drawing objects at random from a given set

of objects, it is termedsampling a population. We need to distinguish between

two different ways in which such asampling experimentmay be performed. After

an object has been drawn at random from the set it may either be put aside

or returned to the set before the next object is randomly drawn. The former is

termed ‘sampling without replacement’, the latter ‘sampling with replacement’.

Find the probability of drawing two aces at random from a pack of cards(i)when the

first card drawn is replaced at random into the pack before the second card is drawn, and

(ii)when the first card is put aside after being drawn.

LetAbe the event that the first card is an ace, andBthe event that the second card is an
ace. Now
Pr(A∩B)=Pr(A)Pr(B|A),

and for both (i) and (ii) we know that Pr(A)= 524 = 131.
(i) If the first card is replaced in the pack before the next is drawn then Pr(B|A)=
Pr(B)= 524 = 131 ,sinceAandBare independent events. We then have

Pr(A∩B)=Pr(A)Pr(B)=

1

13

×

1

13

=

1

169

.

(ii) If the first card is put aside and the second then drawn,AandBare not independent
and Pr(B|A)= 513 , with the result that

Pr(A∩B)=Pr(A)Pr(B|A)=

1

13

×

3

51

=

1

221

.