Mathematical Methods for Physics and Engineering : A Comprehensive Guide

PROBABILITY

where in the last linep=M/Nandq=1−p. This is called thehypergeometric

distribution.

By performing the relevant summations directly, it may be shown that the

hypergeometric distribution has mean

E[X]=n

M

N

=np

and variance

V[X]=

nM(N−M)(N−n)

N^2 (N−1)

=

N−n

N− 1

npq.

In the UK National Lottery each participant chooses six different numbers between 1

and 49. In each weekly draw six numbered winning balls are subsequently drawn. Find the

probabilities that a participant chooses0, 1, 2, 3, 4, 5, 6winning numbers correctly.

The probabilities are given by a hypergeometric distribution withN(the total number of
balls) = 49,M(the number of winning balls drawn) = 6, andn(the number of numbers
chosen by each participant) = 6. Thus, substituting in (30.97), we find

Pr(0) =

(^6) C 043 C 6
(^49) C 6 =

1

2. 29

, Pr(1) =

(^6) C 143 C 5
(^49) C 6 =

1

2. 42

,

Pr(2) =

(^6) C 243 C 4
(^49) C 6 =

1

7. 55

, Pr(3) =

(^6) C 343 C 3
(^49) C 6 =

1

56. 6

,

Pr(4) =

(^6) C 443 C 2
(^49) C 6 =

1

1032

, Pr(5) =

(^6) C 543 C 1
(^49) C 6 =

1

54 200

,

Pr(6) =

(^6) C 643 C 0
(^49) C 6 =

1

13. 98 × 106

.

It can easily be seen that

∑^6

i=0

Pr(i)=0.44 + 0.41 + 0.13 + 0.02 + O(10−^3 )=1,

as expected.

Note that if the number of trials (balls drawn) is small compared withN,M

andN−Mthen not replacing the balls is of little consequence, and we may

approximate the hypergeometric distribution by the binomial distribution (with

p=M/N); this is much easier to evaluate.

30.8.4 The Poisson distribution

We have seen that the binomial distribution describes the number of successful

outcomes in a certain number of trialsn. The Poisson distribution also describes

the probability of obtaining a given number of successes but for situations

in which the number of ‘trials’ cannot be enumerated; rather it describes the

situation in which discrete events occur in a continuum. Typical examples of