Page 47 - Applied statistics and probability for engineers

Section 2-1/Sample Spaces and Events 25

The four white spaces occur between the ive black bars. In the irst step, focus on the bars. The number of permutations

of ive black bars when two are B and three are b is

5!
= 10
2 3!
!
In the second step, consider the white spaces. A code has three narrow spaces w and one wide space W so there are four
possible locations for the wide space. Therefore, the number of possible codes is 10 × 4 = 40. If one code is held back
as a start/stop delimiter, then 39 other characters can be coded by this system (and the name comes from this result).
Combinations
Another counting problem of interest is the number of subsets of r elements that can be selected
from a set of n elements. Here, order is not important. These are called combinations. Every
subset of r elements can be indicated by listing the elements in the set and marking each element
,
with a “ ” if it is to be included in the subset. Therefore, each permutation of r* s and n r− blanks
*
indicates a different subset, and the numbers of these are obtained from Equation 2-3. For exam-
,
,
,
ple, if the set is S = { a b c d}, the subset { , }a c can be indicated as
a b c d
* ∗
Combinations
The number of combinations, subsets of r elements that can be selected from a set of
n
n elements, is denoted as r ( ) or C r n and
⎛ n⎞ n!
n
C r = ⎜ ⎟ = r n r)! (2-4)
−
r ⎝ ⎠
( !

Example 2-13 Printed Circuit Board Layout A printed circuit board has eight different locations in which a

component can be placed. If ive identical components are to be placed on the board, how many
different designs are possible?
Each design is a subset of size ive from the eight locations that are to contain the components. From Equation

2-4, the number of possible designs is
8! =
5 3 ! ! 56

The following example uses the multiplication rule in combination with Equation 2-4 to answer
a more difi cult, but common, question. In random experiments in which items are selected from
a batch, an item may or may not be replaced before the next one is selected. This is referred to
as sampling with or without replacement, respectively.

Example 2-14 Sampling without Replacement A bin of 50 manufactured parts contains 3 defective parts and
47 nondefective parts. A sample of 6 parts is selected from the 50 parts without replacement. That
is, each part can be selected only once, and the sample is a subset of the 50 parts. How many different samples are there
of size 6 that contain exactly 2 defective parts?

A subset containing exactly 2 defective parts can be formed by irst choosing the 2 defective parts from the three
defective parts. Using Equation 2-4, this step can be completed in
⎛ 3⎞ 3!
⎜ ⎟ = 2 1 ! ! = 3 different ways
⎝

2⎠