Mathematical Methods for Physics and Engineering : A Comprehensive Guide

PROBABILITY

In calculating the number of permutations of the various objects we have so

far assumed that the objects are sampledwithout replacement– i.e. once an object

has been drawn from the set it is put aside. As mentioned previously, however,

we may instead replace each object before the next is chosen. The number of

permutations ofkobjects fromnwith replacementmay be calculated very easily

since the first object can be chosen inndifferent ways, as can the second, the

third, etc. Therefore the number of permutations is simplynk. This may also be

viewed as the number of permutations ofkobjects fromnwhere repetitions are

allowed, i.e. each object may be used as often as one likes.

Find the probability that in a group ofkpeople at least two have the same birthday

(ignoring 29 February).

It is simplest to begin by calculating the probability that no two people share a birthday,
as follows. Firstly, we imagine each of thekpeople in turn pointing to their birthday on
a year planner. Thus, we are sampling the 365 days of the year ‘with replacement’ and so
the total number of possible outcomes is (365)k. Now (for the moment) we assume that no
two people share a birthday and imagine the process being repeated, except that as each
person points out their birthday it is crossed off the planner. In this case, we are sampling
the days of the year ‘without replacement’, and so the possible number of outcomes for
which all the birthdays are different is

(^365) Pk= 365!
(365−k)!

.

Hence the probability that all the birthdays are different is

p=

365!

(365−k)! 365k

.

Now using the complement rule (30.11), the probabilityqthat two or more people have
the same birthday is simply

q=1−p=1−

365!

(365−k)! 365k

.

This expression may be conveniently evaluated using Stirling’s approximation forn!when
nis large, namely

n!∼

√

2 πn

(n

e

)n

,

to give

q≈ 1 −e−k

(

365

365 −k

) 365 −k+0. 5

.

It is interesting to note that ifk= 23 the probability is a little greater than a half that
at least two people have the same birthday, and ifk= 50 the probability rises to 0.970.
This can prove a good bet at a party of non-mathematicians!

So far we have assumed that allnobjects are different (ordistinguishable). Let

us now considernobjects of whichn 1 are identical and of type 1,n 2 are identical

and of type 2,...,nmare identical and of typem(clearlyn=n 1 +n 2 +···+nm).

From (30.30) the number of permutations of thesenobjects is againn!. However,