Download Counting Principles (combinations and

Counting Principles
(Permutations and Combinations )
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Example
A combination lock can be set to open to any 4-digit
sequence.
(a) How many sequences are possible?
(b) How many sequences are possible if no digit is repeated?
Solution: (a) Since there are 10 digits namely 0, 1, 2…..9, there
are 10 choices for each of the digit. By the multiplication
principle, there are 10 ∙10 ∙10 ∙10 =10,000 different sequences.
(b)There are 10 choices for the first digit. It cannot be used
again, so there are 9 choices for the second digit, 8 choices for
the third digit, and then 7 choices for the fourth digit.
Consequently, the number of such sequences is
10 ∙ 9 ∙ 8 ∙7 =5040 different sequences.
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Example
How many different ways can you choose a bagel,
muffin or donut to eat and coffee or juice to drink?
Tree diagram
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Example
A teacher is lining up 8 students for a spelling bee. How many
different line-ups are possible?
Solution: Eight choices will be made, one for each space that will
hold a student. Any of the students could be chosen for the first
space. There are 7 choices for the second space, since 1 student
has already been placed in the first space; there are 6 choices for
the third space, and so on.
By the multiplication principle, the number of different possible
arrangements is 8 ∙7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙1 = 40,320.
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(order does matter)
TI-83/84
function
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Example
A teacher wishes to place 5 out of 8 different books on her
shelf. How many arrangements of 5 books are possible?
Solution : The teacher has 8 ways to fill the first space, 7 ways
to fill the second space, 6 ways to fill the third, and so on…
Since the teacher wants to use only 5 books, only 5 spaces can
be filled (5 events) instead of 8, for 8 ∙7 ∙ 6 ∙ 5 ∙ 4 = 6720
arrangements.
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Example
Find the number of permutations of the letters L, M, N, O,
P,
and Q, if just three of the letters are to be used.
Solution: P(n, r )  n! . Here n  6 and r  3.
(n - r )!
6!
6!


(6 - 3)! 3!
6  5  4  3  2 1

 120
3  2 1
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How many permutations are there of two letters from the set
{A,B,C}.
n!
P(n, r ) 
(n  r )!
3!
3!
P(3,2) 
  3 2  6
(3  2)! 1!
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(order does not matter)
TI-83/84
function
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Example
How many committees of 4 people can be formed
from a group of 10 people?
Solution: A committee is an unordered group, so use the
combinations formula for C(10,4).
10!
10!
C (10, 4) 

4!(10  4)! 4!6!
10  9  8  7  6  5  4  3  2 1 10  9  8  7


 210
4  3  2  1  6  5  4  3  2 1 4  3  2  1
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© 2012 Pearson Education, Inc.. All rights reserved.
Example
From a class of 15 students, a group of 3 or 4 students will be
selected to work on a special project. In how many ways can a
group of 3 or 4 students be selected?
Solution: The number of ways to select group of 3 students
from a class of 15 students is C(15, 3) = 455.
The number of ways to select group of 4 students from a class
of 15 students is C(15, 4) = 1365.
The total number of ways to select a group of 3 or 4 students
will be the 1820.
© 2012 Pearson Education, Inc.. All rights reserved.
© 2012 Pearson Education, Inc.. All rights reserved.
Example
(a) How many 4-digit code numbers are possible if no digits are
repeated?
Solution: Since changing the order of the 4 digits results in a
different code, permutations should be used.
(b) A sample of 3 light bulbs is randomly selected from a batch
of 15. How many different samples are possible?
Solution: The order in which the 3 light bulbs are selected is not
important. The sample is unchanged if the items are rearranged,
so combinations should be used.
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Example
(c) In a baseball conference with 8 teams, how many games
must be played so that each team plays every other team exactly
once?
Solution: Selection of 2 teams for a game is an unordered
subset of 2 from the set of 8 teams. Use combinations again.
(d) In how many ways can 4 patients be assigned to 6 different
hospital rooms so that each patient has a private room?
Solution: The room assignments are an ordered selection of 4
rooms from the 6 rooms. Exchanging the rooms of any 2
patients within a selection of 4 rooms gives a different
assignment, so permutations should be used.
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Introduction to Probability
Set – a collection of elements that satisfy a certain condition.
Write the elements belonging to the set {x | x is a state whose
name begins with the letter O}.
Solution: The state names that begin with the letter O make up
the set {Ohio, Oklahoma, Oregon }.
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Example 1
Decide if the statement is true or false.
{2, 4,6}  {6, 2, 4}
Solution: The first set is a subset of the second because
each element of first set belongs to second set. (The fact
that the elements are listed in a different order does not
matter.) Therefore, the statement is true.
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© 2012 Pearson Education, Inc.. All rights reserved.
Sample Space – The set of all possible outcomes of an
experiment or event.
A) What is the sample space?
B) What is the probability of getting a 3?
C) What is the probability of getting an
odd number?
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Example
Two coins are tossed, and a head or a tail is recorded for each
coin. Write the event E: the coins show exactly one head.
Solution: Tossing a coin is made up of the outcomes heads (H)
or tails (T). If S represents the sample space of tossing two
coins, then S = {HH, HT, TH, TT}.
Two outcomes satisfy this condition: so,
E = {HT, TH}.
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Number of elements
in the sets
Example
Two coins are tossed, and a head or a tail is recorded for each
coin.
A)
Give a sample space for this experiment.
B)
What is the probability of getting at least one head.
Solution: A) Tossing a coin is made up of the outcomes heads (H)
or tails (T). If S represents the sample space of tossing two
coins then S = {HH, HT, TH, TT}.
Solution : B) E = {HH,HT,TH}
P( E ) 
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n( E ) 3

n( S ) 4
 0.152
 0.384
 0.56
 0.616
 0.179
 0.0625
 0.625
 0.25
 0.75
Monty Hall probability puzzle
http://www.youtube.com/watch?v=mhlc7peGlGg&safe=active
 0.824
 0.712
Probability of Multiple Events
(E’ is called the complement of E)
(no outcomes in common; if one event occurs the other cannot)
4
1
6
5
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EF  6
2
Example
A study of workers earning the minimum wage grouped such
workers into various categories, which can be interpreted as
events when a worker is selected at random. Source:
Economic Policy Institute. Consider the following events:
E: worker is under 20;
F: worker is white;
G: worker is female.
Describe the following event in words:
Solution:
E  F .
E  F  is the event that the worker is not under 20
and the worker is not white.
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Independent Events
The fact that one event has occurred, does not change the probability
of the second event occurring.
** Events that are mutually exclusive must be dependent, but
dependent events are not necessarily mutually exclusive**
Read as; the
probability of event
F occurring given
E has occurred
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Read as; the
probability of event
E occurring given
F has occurred
(General rule of addition)
P( E  F )  0
So the equation simplifies to P( E  F )  P( E )  P( F )
For mutually exclusive events,
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Example
If a single card is drawn from an ordinary deck of cards, find
the probability of an ace or a club.
Solution: Let A represent the event “an ace ” and C the event
“club card.” There are 4 aces in the deck, so P( A)  4 / 52.
There are 13 clubs in the deck, so
P(C )  13 / 52.
Since there is 1 ace of club in the deck, P ( A C )  1/ 52.
By the union rule, the probability of the card being an ace or a
Club card is
P( A C )  P( A)  P(C )  P( A C ).
4 13 1
16 4
  
  .
52 52 52
52 13
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Example
Suppose two fair dice are rolled. Find the probability that the sum is 8, or both
die show the same number.
Solution:
1-1
1-2
1-3
1-4
1-5
1-6
2-1
2-2
2-3
2-4
2-5
2-6
3-1
3-2
3-3
3-4
3-5
3-6
4-1
4-2
4-3
4-4
4-5
4-6
5-1
5-2
5-3
5-4
5-5
5-6
6-1
6-2
6-3
6-4
6-5
6-6
By the union rule, P(sum is 8 or both die show same number)
5 6 1 10 5
=   
 .
36 36 36 36 18
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Example
Find the probability that when two fair dice are rolled, the sum
is less than 11.
Solution: To calculate this probability directly, we must find the
probabilities that the sum is 2, 3, 4, 5, 6, 7, 8, 9 or 10 and then
add them. It is much simpler to first find the probability of the
complement, the event that the sum is greater than or equal to 11.
P(sum  11)  1  P(sum  11)
3 33 11
 1   .
36 36 12
1-1
1-2
1-3
1-4
1-5
1-6
2-1
2-2
2-3
2-4
2-5
2-6
3-1
3-2
3-3
3-4
3-5
3-6
4-1
4-2
4-3
4-4
4-5
4-6
5-1
5-2
5-3
5-4
5-5
5-6
6-1
6-2
6-3
6-4
6-5
6-6
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(General rule of multiplication)
When events are Independent;
P( E | F )  P( E )
So the equation simplifies to:
P( F | E )  P( F )
Independent
Not
Independent
Not
Independent
Independent
1/6
9/34
0.54
2/7
9/25
Not mutually exclusive
mutually exclusive
Not mutually exclusive
P(F); represented by large
circle.
P(E and F); represented by
overlapping section
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Example
The following table shows national employment
statistics. Use the table to find
each probability.
9729
9392
8937
5801
4383
19121
7. P(male | professional)
4190
 0.47
8937
10. P(professional | female)
4747
 0.51
9392
8. P(laborer | female)
1432
 0.15
9392
11. P(sales | male)
2588
 0.27
9729
9. P(female | sales)
3213
 0.55
5801
12. P(male | laborer)
2951
 0.67
4383
Example
Use a tree diagram to solve the following problem.
16. A car insurance company compiled the following information from a recent
survey. 75% of drivers carefully follow the speed limit Of the drivers who carefully
follow the speed limit, 80% have never had an accident. Of the drivers who do not
carefully follow the speed limit, 65% have never had an accident.
What is the probability that a driver does not carefully follow the speed limit and has
never had an accident?
C is the event careful
driver.
0.80
NA is the event no
accident
0.75
0.20
0.65
P (C  NA)  0.6
P(C  NA' )  0.15
P(C 'NA)  0.1625
0.25
0.35
P(C 'NA' )  0.0875