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Chapter 2. Foundations of Probability
Section 2.4. Counting Rules Useful in Probability
Jiaping Wang
Department of Mathematical Science
01/23/2013, Wednesday
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Outline
Fundamental Principle of Counting
Permutations
Combinations
Partitions
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Part 1. Fundamental Principle of
Counting
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Examples
A relative frequency definition of probability will work for any experiment that results
a finite sample space with equally likely outcomes.
So counting becomes a key step in obtaining the probability.
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Theorem 2.2
Fundamental Principle of Counting:
If the first task of an experiment can result in n1 possible
outcomes and for each such outcome, the second task can
result in n2 possible outcomes, then there are n1n2 possible
outcomes for the two tasks together.
The principle can extend to more tasks in a sequence.
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Example 2.5
In connection with the national retail chain that plans to
build two new stores, the eight possible combinations
of locations are as shown in the figure. If all eight
choices are equally likely (that is, if one of the pairs of
cities is selected at random), find the probability that
City E is selected.
Solution: As E can appear in the combinations
AE, BE, CE or DE, also there are total 8 possible
combinations with equal chance to be selected,
thus the probability of City E is selected is 1/8*4=1/2.
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Example 2.6
Five cans of paint (numbered 1 through 5)
were delivered to a professional
painter. Unknown to her, some of the
cans (1 and 2) are satin finish and the
remaining cans (3, 4 and 5) are glossy
finish. Suppose she selects two cans at
random for a particular job. Let A
denote the event that the painter
selects the two can of statin-finish paint
and let B denote the event that the two
cans have different finishes (one of
satin and one of glossy). Find P(A) and
P(B).
Solution: Total we have 20 possible combinations.
For event A, there are two possibilities: {1,2} or {2,1},
so P(A)=2/20;
For event B, there are 12 possibilities: {1,3}, {1,4},
{1,5}, {2,3}, {2,4}, {2,5}, {3,1}, {3,2}, {4,1}, {4,2}, {5,1},
{5,2}, so P(B)=12/20.
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Part 2. Permutations
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With/Without Replacement
Another Example: select telephone numbers. For example, given the area
code (940), then how many possibilities can one choose from 0-9
numbers in the next 7 digits?
Answer: the number of possible first digit is 10, the number of possible 2nd
digit is also 10, and so for other digits. So total we have
10x10x10x10x10x10x10=107.
From this example and Example 2.6, we can find two things affect
the manner of counting:
First is whether or not the order is important;
Second is whether or not chosen with replacement or without replacement.
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Permutations
Theorem 2.3 Permutations. The number of ordered arrangements or
permutations Prn of r objects selected from n distinct objects (r≤n) is
given by
n! = n×(n-1) ×(n-2) ו••×3×2×1, 0!=1, n!=n(n-1)!.
Order Is Important
With Replacement
nr
Without Replacement
Prn
Order Is Not Important
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Example 2.7
A small company has 12 account managers. Three potential
customers have been identified and each customer has
quite different needs. The company’s director decides to
send an account manager to visit each of the potential
customers and considers the customer’s needs in making
his selection. How many ways are there for him to assign
three different account managers to make the contacts?
Solution: P312 = 12!/(12-3)!=12!/9!=12 x 11 x 10 x 9!/9!=1320.
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Part 3. Combinations
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Order Is Not Important
For example, playing bridge with dealing 13
cards, at this time, the order in which the
cards are dealt does not affect the final hand.
For this, the order is not important; Also, this
selection is without replacement. For this
kind of case, it is called combination.
Another example in counting
based on combination is drawing
the numbers in lottery.
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Combinations
Theorem 2.4 Combinations. The number of distinct subsets
or combinations of size r that can be selected from n distinct
objects (r ≤ n)is given by
Order Is Important
With Replacement
nr
Without Replacement
Prn
Order Is Not Important
Crn
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Example 2.8
Most states conduct lotteries as a means of raising
revenue. In Florida’s lottery, a player select 6 numbers
from 1 to 53. For each drawing, balls numbered from 1 to
53 are placed in a hopper. Six balls are drawn from the
hopper at random and without replacement. To win the
jackpot, all six of the player’s number must match those
drawn in any order. How many wining numbers are
possible?
Solution: C653 = 53!/[6! (53 -6)!]=53x52x51x50x49x48/(6x5x4x3x2x1)=22957480
So the probability to win is about 1/22957480.
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Example 2.9
Twenty six states participate in the Powerball lottery. In this lottery, a player
selects five number between 1 and 53 and a Powerball number between 1
and 42. For each drawing, five balls are drawn at random and without
replacement from a hopper with 53 white balls numbered 1 to 53. A sixth ball
is drawn from a second hopper with 42 red balls numbered 1 to 42. To win
the jackpot, the five numbers selected by the player must match those of the
five white balls drawn, and the player’s Powerball number must match the
number on the red ball drawn from the hopper. How many possible wining
numbers are there? Is there a greater probability of wining Florida’s lottery or
the Powerball if one buys a single ticket?
Solution: there are two tasks, in the first task, the number is
C5 53 = 53x52x51x50x49/(5x4x3x2x1)=2869685;
in the second task, there are C142 = 42.
Based on the Fundamental Principle of Counting, there are total
C5 53 x C142 = 2869685 x 42 = 120,536,770, then we can say the probability of
winning the Powerball lottery is 1/ 120,536,770, which is less than the probability
of winning Florida’s lottery.
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Example 2.10
A department in a company has 12 members: 8 males and 4 females.
To gain greater insight into the employees’ view of various benefits,
the human resources office plans to form a focus group from
members of this department. Five members will be selected at
random from the department’s members. What is the probability that
the focus group will only have males? What is the probability that the
focus group will have two males and three females?
Solution: It is selection without replacement and order is not important. So the total
number of ways is 12!/(5!7!)=792;
Q1: The number of ways to select only males is that select 5 males from 8 which
is 8!/(5!3!)=56, so the probability is 56/792.
Q2: The number of ways to select 2 males is 8!/(2!6!)=28 and the number of ways
to select 3 females is 4!/(3!1!)=4, then the probability is 28x4/792.
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Part 4. Partitions
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Theorem 2.5 Partitions
Consider a case: If we roll a die for 12 times, how many possible ways to have
2 1’s, 2 2’s, 3 3’s, 2 4’s, 2 5’s and 1 6’s?
Solution: First, choose 2 1’s from 12 which gives 12!/(2!10!), second, since there
two positions are filled by 1’s, the next choice appears in the left 10 positions, so
there are 10!/(8!2!) ways, and so similar for next other selections which provides
final result is 12!/(2!10!)x10!/(2!8!)x8!/(3!5!)x5!/(2!3!)x3!/(2!1!)x1!/(1!0!)
=12!/(2!x2!x3!x2!x2!x1!)
Theorem 2.5 Partitions. The number of partitioning n
distinct objects into k groups containing n1, n2,•••, nk
objects, respectively, is
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Examples 2.11, 2.12
2.11 Suppose 10 employees are to be divided among three job
assignments, with 3 employees going to job I, 4 to job II, 3 to job III. In
how many ways can the job assignments be made?
Solution: 10!/(3!4!3!)=4200.
2.12 Suppose that three employees of a certain ethnic group all get
assigned to job I. Assumes that they are the only employees among
the 10 under consideration who belong to this ethnic group, what is
the probability of this happening under a random assignment of
employees to jobs?
Solution: As the job I is filled by these three employees with the same ethnic, we
only consider the left employees. At this time, we have total 7 employees with
two different groups (job II and job III), so the number of ways is 7!/(4!3!)=35, then
we can find the probability is 35/4200 = 1/120.
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Determining whether or order is important
Is critical to counting
Suppose a fair coin is flipped three times and the number of heads
observed.
Unordered
TTT
HTT
HHT
HHH
Ordered
TTT
HTT, THT, TTH
HHT, HTH, THH
HHH
Probabilities
1/8
3/8
3/8
1/8
From here, we can find the probabilities are same for ordered or unordered
results, but the sample space is changed.
A key to determining whether or not order is important is an understanding of the
random process giving rise to the outcomes.
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Example 2.13
The birthday problem. Suppose n people are in a room. Assume no one has
birthday at 2/29 and there are total 365 days per year.
Q1: What is the probability that no two of them have the same birthday?
Solution: Each person can have 365 possible birthdays, so there are total 365n ways.
If for the first person, there are 365 possible birthdays, then the second one has only
364 possible birthdays (as the first one filled one of them) and so on for the next
Persons), so total there are 365(364)••(365-n+1) ways such that no two have the same
Birthday. Then we have the probability 365(364)••(365-n+1)/ 365n . If n>365, based on the
Pigeon Hole theorem, at least two of them have the same birthday, so the probability that no
two of them have the same birthday is 0.
Q2: How many people must be in the room for the probability that at least
two of the n people have the same birthday to be greater than 1/2?
Solution: Consider the opposite event: no two of them have the same birthday, so
the probability is 365(364)••(365-n+1)/ 365n <1/2, by solving this inequality to obtain
The maximum n=22, so the opposite event should need 23 people.
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Example 2.14
A poker hand consist of five cards. If all of them are from the same
suit but not in consecutive order, we say that the hand is a flush. For
instance, if we have five clubs that are not in consecutive order (say,
2, 4, 5, 6, 10), then we have a flush. What is the probability of a flush
but not straight flush?
Solution: There are total 52!/(47!5!) ways to choose 5 from 52 cards. Now to cho
The flush, first there are 4 different suits, second given a suit, there are 13!/(5!8!
including the straight flush. Now we know there are 10 different straight flush fro
(ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, ace). So the probability would be
4x(13!/(5!8!)-10)/(52!/(47!5!).
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Homework #2
Page 45: 2.36, 2.38, 2.40, 2.42, 2.43,
Page 46: 2.46, 2.48, 2.50
Due Wed., 03/02/2013
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Part 5. More Counting Rules
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Order and Replacement
Theorem 2.6 The number of ways of making r selections
from n objects when selection is made with replacement
and order is not important is
Order Is Important
Order Is Not Important
With Replacement
nr
Crn+r-1
Without Replacement
Prn
Crn
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Example 2.16
In a lottery, a player selects 6 numbers between 1 and 44.
The same number may be chosen more than once. For
each drawing, 6 balls are drawn at random with
replacement from a hopper with 44 white balls numbered
1 to 44. Sufficient time is allowed between selections of a
ball for the previously selected ball to be mixed with the
others. To win the jackpot, all six numbers of the player
must match those drawn in any order. How many wining
numbers are possible?
Solution: C644+6-1 = 49!/(6!43!)
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