Tải bản đầy đủ
Appendix 5a. Counting Rules for Computing Probabilities

# Appendix 5a. Counting Rules for Computing Probabilities

Tải bản đầy đủ

RANDOM VARIABLES AND PROBABILIT Y DISTRIBUTIONS

247

Product Rule
RULE 1: THE PRODUCT RULE
Suppose there are r groups of objects. In group 1 there are n1 objects, in group
2 there are n2 objects, and so on. If we define the experiment to be the selection
of one object from each of the r groups, then there are n1 ⋅n2 ⋅. . .⋅nr elementary
objects in the experiment.
To illustrate the use of this counting rule, consider the following example.
EXAMPLE 5a.1. A planner is interested in surveying retail patronage at three types
of shoe stores: factory outlets, discount stores, and specialty stores. She decides to
sample one store of each type from the outlets in the town. If there are 10 factory outlets, 7 discount stores, and 3 specialty stores in the town, how many different combinations of stores are there?
There are r = 3 groups of objects, with n1 = 10, n2 = 7, and n3 = 3. If one store
must be selected from each category, then there are (10)(7)(3) = 210 possible combinations of stores from which the planner may choose.

Combinations Rule
The second and third counting rules are concerned with a slightly different problem.
The experiment consists of selecting a subset of r objects from a set of n objects,
where r ≤ n.
EXAMPLE 5a.2. Suppose a city is considering adding two new libraries to its existing system. To minimize expenses, the libraries will be constructed on two of five
vacant properties currently owned by the city. How many different locational plans
for the system of libraries are there?
Label the five available sites as A, B, C, D, and E. The possible outcomes of the
experiment are:
AB
AC
AE

BC
BD
BE

CD
CE

DE

Note that each of the possible plans contains two different sites; these are combinations. Note also that we list CD as a site combination, but not DC. The reason is that
by definition, order does not matter in a combination. From the city’s perspective,
there is no difference between outcome CD and DC, because buildings will be erected
on the same two sites.

248

I N F E R E N T I A L S TAT I S T I C S

DEFINITION: COMBINATION
A combination is a set of r distinguishable objects, regardless of order.
The counting rule for combinations is concerned with the number of different combinations of size r that can be drawn from a master set of n objects.
RULE 2: COMBINATIONS RULE
The number of combinations of r objects taken from n different objects is
n!
Crn = ————
r!(n – r)!
Sometimes the number of combinations Crn is written C(n, r) or ( rn ). The symbol n! is
read “n factorial” and is defined as n ⋅ (n – 1) ⋅ (n – 2) ⋅ . . . ⋅ 2 ⋅1. For example, 5! =
5 ⋅4 ⋅ 3⋅2 ⋅ 1 = 120 and 2! = 2 ⋅1 = 2. By definition, 0! = 1.
For the library example, n = 5 and r = 2; therefore
5!
5 ⋅ 4 ⋅ 3! 20
C25 = ———— = ——— = —– = 10
2!(5 – 2)! 2 ⋅ 1 ⋅3! 2
These are the 10 combinations listed above. As you can imagine, the combinations
rule can save a great deal of time in enumerating the elementary outcomes of an
experiment.

Permutations Rule
Let us now consider a slightly different problem. Suppose the city has five sites for
two new libraries, one that will house the main library and one that will house digital and other special collections. The sites are labeled A to E as before. Now consider
the outcome AB. This is interpreted as meaning site A is to be used for the main library (the first letter of the pair) and site B (the second letter) is to be used for the special collections. Notice that outcome AB is not the same as outcome BA, because the
two libraries are placed on different sites. In BA, the main library is located at site B
while in AB the main library is on site A. In this instance, the order of the elements in
the set is important.
DEFINITION: PERMUTATION
A permutation is a distinct ordering of r objects.
The counting rule for permutations defines the number of different ordered arrangements of size r that can be drawn from a set of n objects.

RANDOM VARIABLES AND PROBABILIT Y DISTRIBUTIONS

249

RULE 3: PERMUTATIONS RULE
The number of permutations of r objects taken from a set of n different objects is
n!
Prn = ———
(n – r)!
Sometimes, Prn is written as P(n, r). In the library example, n = 5 and r = 2, so that
5!
P25 = ——— = 20
(5 – 2)!
That is, there are twice as many permutations as combinations in a problem of this
size. For each combination AB there is the reverse arrangement BA. This is not a general result. The formal relation between the number of combinations and the number
of permutations is given by
1
Crn = — ⋅ Prn
r!
Whenever r = 2, there are twice as many permutations as combinations.

Hypergeometric Rule
The final rule combines the product rule with the combinations rule.
RULE 4: HYPERGEOMETRIC RULE
A set of n objects consists of n1 of type 1 and n2 of type 2, where n1 + n2 = n.
From this group of n objects, define an experiment in which r1 of the first type
and r2 of the second type are to be chosen. The number of different groups that
can be drawn is
Cr1n1 ⋅ Cr2n2
As an example, consider the following problem. A survey is to be undertaken
of neighborhoods in some city. There are 10 suburban neighborhoods and 8 inner-city
neighborhoods. A decision is to be made to survey three suburban and two inner-city
neighborhoods. How many different combinations of neighborhoods are possible for
the survey? Clearly, the hypergeometric rule applies. There are two groups with n1 =
10 and n2 = 8. From these two groups, r1 = 3 and r2 = 2 objects are to be drawn. From
the hypergeometric rule, the number of combinations of neighborhoods that could be
drawn is
10!
8!
C310 ⋅ C28 = —–——— ⋅ ———— = 120(28) = 3360
3!(10 − 3)! 2!(8 − 2)!

250

I N F E R E N T I A L S TAT I S T I C S

Appendix 5b. Expected Value and Variance of a Continuous
Random Variable
There is a direct correspondence between the area under a probability density function
between any two limits c and d and the probability that a continuous random variable
X assumes a value between these two limits. Students familiar with calculus will recognize that the area under f(x) between c and d is given by

∫c f(x)dx
d

where the symbol ∫ represents the integration operator. It is directly analogous to the
summation operator Σ used for discrete variables. The requirement that the sum of
the probabilities of the values that may be assumed by a random variable equal one
can be stated as

∫a f(x)dx = 1
b

where a and b are the limits of the values that may be assumed by the random variable.
For continuous random variables defined over the infinite length of the real number
line, the appropriate equation is

∫ −∞ f (x)dx = 1
A second important property is that the probability of a continuous variable assuming
a particular value, say X = c, is always zero. That is,

∫c f(x)dx = 0
c

because any definite integral over the limits c to c must be zero.
The expected value of a continuous random variable is
E(X) = ∫ xf(x)dx
b

a

and the variance is
V(X) = ∫ [x – E(X)]2 f(x)dx = ∫ x2 f (x)dx – [E(X)]2
b

b

a

a

As an example, let us consider the case of the uniform continuous random variable
defined by

{

1
—–
f (x) = 10
0

0 ≤ x ≤ 10
otherwise

RANDOM VARIABLES AND PROBABILIT Y DISTRIBUTIONS

251

First, to show the area under f(x) equals one, we note

∫0

10

1
1
—– dx = —– (10 – 0) = 1
10
10

To calculate the expected value, we write

( )

1
1 10
10
E(X) = ∫ x —– dx = —– ∫ xdx
0
10
10 0
1 x2 10
= —– —–⏐
10 2 0
1 102 02
= —– —– – — = 5
10 2
2

(

)

The variance is

( )

1
10
V(X) = ∫ x2 —– dx – [E(X)]2
0
10
1 x3 10
= —– —–⏐ – 52
10 3 0

(

)

1 1000
= —– ——– – 25 = 8.33
10
3