Transcript Week 1: Descriptive Statistics

Topic 3: Counting
Techniques
CEE 11 Spring 2002
Dr. Amelia Regan
These notes draw liberally from the class text, Probability and Statistics for
Engineering and the Sciences by Jay L. Devore, Duxbury 1995 (4th edition)
Counting Techniques

When the various outcomes of an experiment are equally likely
then the task of computing probabilities reduces to counting. In
particular, if N is the number of outcomes in the sample space
and N(A) is the number of outcomes contained in an event A,
then
P ( A) 

N ( A)
N
The product rule for ordered pairs:
If the first element or object of an ordered pair can be selected in
n1 ways and for each of these n1 ways the second element of the
pair can be selected in n2 ways, then the number of pairs is n1n2
Review Questions

Problem #57, chapter 2
 For any two events A and B with P(B) > 0
Show that P(A|B) + P(A’|B) = 1.0
Do this mathematically or with a venn diagram
Review Question

Solution
P( A  B) P( A ' B)

P( B)
P( B)
P( B  A) P( B  A ') P( B  A) P( B | A) P( A)  P( B | A ') P( A ') P( B)




P( B)
P( B)
P( B)
P( B)
P( B)
P( A | B)  P( A ' | B) 

Note -- we use the law of total probability
P( B)  P( B | A1 ) P( A1 )  P( B | A2 ) P( A2 )  ...  P( B | An ) P( An )
n
  P( B | Ai ) P( Ai )
i 1
Review Questions

Problem #58, chapter 2
 Show that for any three events A, B and C with P(C) > 0

(AUB|C) = P(A|C)+P(B|C)-P(A intersection B|C)
Counting Techniques

For small problems, we sometimes construct a tree diagram to
enumerate the different possible combinations. Assume that a
UCI student must choose from three dorms and that he or she
must select a room with zero, one or two roommates. The tree
diagram below shows the nine possible combinations.
D1
D2
D3
r0
r1
r2
r0
r1
r2
r0
r1
r2
Permutations



The number of permutations of n objects is equal to n(n-1)(n-2)…(2)(1) or n!
For example, if a traveling salesman must visit 10 customers on a particular
day there are 10! different routes that he may take.
Any ordered sequence of k objects taken from a set of n distinct objects is
called a permutation of size k of the objects. The number of permutations of
size k that can be constructed from the n objects is denoted by Pk,n
The number of permutations of size k that can be constructed from n objects
is equal to n(n-1)(n-2)…(n-k+1)
Pk , n  n(n  1),...(n  k  1)
n(n  1),..., (n  k  1)(n  k )(n  k  1)
(n  k )(n  k  1),...(2)(1)
n!

(n  k )!

Permutations

Sometimes we need to determine the number of
permutations possible when some of the objects in a
group are indistinguishable from each other.
For example: (Ross, 1988, p5)
How many different letter arrangements can be formed
using the letters PEPPER?
Permutations
Solution: First note that there are 6! Permutations of the letters
P1 E1 P2 P3 E2 R when the 3 P’s and 2 E’s are distinguished from
each other.
However consider any one of these permutations -- for instance, P1
P2 E1 P3 E2 R.
If we now permute the P’s among themselves and the E’s among
themselves then the resultant arrangement would still be of the
form PPEPER. That is, all 3!2! Permutations are of the form
PPEPER. Hence there are 6!/3!2! = 60 possible letter
arrangements of the letters PEPPER.
combinations


We are often interested in determining the
number of different groups of k objects may be
formed from a group of n objects
Given a set of n distinct objects, any unordered
subset of size k of the objects is called a
combination. The number of combinations of size
k that can be formed from n distinct objects is
denoted by  n  or sometimes by Ck,n
k
 n  Pk ,n
n!


 
k!
k !(n  k )!
k
Class exercise

To successfully route luggage from one airport to
another the destination airport is identified by
three letters. For example, Los Angeles is LAX,
Orange County is SNA. How many airports can
be served by this identification system?
Class exercise



A student looking for a good used car finds ten
cars in his price range advertised on the web.
Since he has to ride the bus to look at the cars in
person he has only enough time to look at four of
them.
Order matters in this case because if he finds a
car he likes at any time in the process he’ll buy it.
In how many ways can the four cars be selected
from the set of ten?
If now we assume that he’ll look at all four cars
and then make his decision (order does not
matter). In how many ways can the four cars be
selected from the set of ten?
Class exercise

Using the principle that
P ( A) 

N ( A)
N
What is the probability of dealing a perfect bridge
hand? (13 cards of the same suit?)