Counting and Probability
•Sets and Counting•Permutations & Combinations•Probability
A set is a well-defined collection of distinct objects.
Well-defined means there is a rule that enables us to determine whether a given object is an element of the set.If a set has no elements, it is called the empty set, or null set, and is denoted by the symbol .
Sets and Counting
If two sets A and B have precisely the same elements, then A and B are said to be equal and write A = B.
If each element of a set is also an element of
a set , then we say that is a subset of and
write
A
B A B
A B
If and , then we say that is a
of and write
A B A B A
B A B
proper subset .
A A for any set
Write down all subsets of x y z, , .
0 elements:
1 element: x y z, ,
2 elements: x y x z y z, , , , ,
3 elements: x y z, ,
If and are sets, the of with ,
denoted, , is the set consisting of elements
that belong to both and . The of
with , denoted , is the set consisting of
elements that belong to or , or both.
A B A B
A B
A B A
B A B
either A B
intersection
union
Designate the denoted
as the set consisting of all the elements
we wish to consider.
universal set, ,U
If is a set, the of , denoted ,
is the set consisting of all the elements in the
universal set not in .
A A A
A
complement
U
A
B
A B
U
BA
A B
U
BA
A B
U
B
A B
A
U
AA
A
A
U
A B
A B A B A
blue, red, green, yellow, orange
blue, red blue, green, orange
Find
(a) (b) (c)
(a) blue, red, green, orangeA B
(b) blueA B
(c) green, yellow, orangeA
Theorem: Counting Formula
If A and B are finite sets, then
n A B n A n B n A B
In survey of 50 people, 21 said they owned stocks, 32 said they owned bonds and 12 said they owned both stocks and bonds. How many of the 50 people owned stocks or bonds? How many owned neither?
A: person owns stock B: person owns bonds
n A B n A n B n A B( ) ( ) ( ) ( )
= 21 + 32 - 12 = 41
50 - 41 = 9 owned neither
Universe is 50 people. In A = 21 owned stocks. In B = 32 owned bonds. In AB = 12 owned both stocks and bonds. In AB = 53-12 = 41 owned stocks or bonds. In (AB)= 50-41 = 9 owned neither.
A 21
B 32
(AB) _ _ 9
AB _12
Permutations and Combinations
This is a tricky subject where even the text author makes mistakes. The next three slides are to distinguish some of the subtleties. After these are the slides from the text set and that help to elaborate some cases. Some situations can be very difficult to evaluate.
Counting Permutation Cases I A permutation is an ordered arrangement of r
objects from n objects. To find the total number of possible cases, the easy types of situations are:
1. Multiplication of p,q,r,s, ... ways of selection => Total number of cases is N = p•q•r•s•...
2. The number of permutations of r distinct objects with allowed repetition from n distinct objects (order is important) => N = nr
3. The number of permutations of r distinct objects with no repetition from n distinct objects (order is important) => N = P(n, r) = n!/(n - r)!
Counting Permutation Cases II A permutation is an ordered arrangement of r
objects from n objects.
An important harder situation of finding the total number of possible cases is when there are k distinct types each of non-distinct objects, with n1 of the 1st type, n2 of the 2nd type, ... nk of the kth type, and n = n1 + n2 + n3 + ... nk.
4. Permutation of n, some non-distinct, objects with allowed repetition from k distinct types of objects (order is important) =>
N = n!/[n1! • n2! ... • nk!].
Counting Combination Cases A combination is an arrangement with no
regard to order of r distinct objects without repetition from n distinct objects (r < n). For finding the total number of possible cases, the easy type of situation is:
The number of combinations of r distinct objects without repetition from n distinct objects (order is not important) =>
N = C(n, r) = n!/[r!(n - r)!].
Theorem: Multiplication Principle of Counting
If a task consists of a sequence of choices in which there are p selections for the first choice, q selections for the second choice, r selections for the third choice, and so on, then the task of making these selections can be done in
p q r different ways.
Permutations and Combinations
If a license plate consists of a letter, then 5 numbers, how many different types of license plates are possible?
26 10 10 10 10 10 2 600 000 , , license plates
A permutation is an ordered arrangement of n distinct objects without repetitions. The symbol P(n, r) represents the number of permutations of n distinct objects, taken r at a time, where r < n.
Theorem: Number of Permutations of n Distinct Objects Taken r at a Time
The number of different arrangements from selecting r objects from a set of n objects (r < n), in which
1. the n objects are distinct2. once an object is used, it cannot be repeated
3. order is importantis given by the formula
P n rn
n r( , )
!( )!
Evaluate: P( , )10 3
P( , )!
( )!10 3
1010 3
107
!!
10 9 8 77
!!
720
A combination is an arrangement, without regard to order, of n distinct objects without repetitions. The symbol C(n, r) represents the number of combinations of n distinct objects taken r at a time, where r < n.
Theorem: Number of Combinations of n Distinct Objects Taken r at a Time
The number of different arrangements from selecting r objects from a set of n objects (r < n), in which
1. the n objects are distinct
2. once an object is used, it cannot be repeated
3. order is not important
is given by the formula
.)!(!
!)(
rnr
n
r
nCrn,C rn
Evaluate: 3)C( ,10
C( , )!
!( )!10 3
103 10 3
103 7
!! !
10 9 8 73 7
!! !
10 9 83 2 1
120
An event is an outcome from an experiment. Its probability gives the likelihood it occurs.A probability model lists the different outcomes from an experiment and their corresponding probabilities.
To construct probability models, we need to know the sample space of the experiment. This is the set S that lists all the possible outcomes of the experiment.
Probability
Determine the sample space resulting from the experiment of rolling a die.
S = {1, 2, 3, 4, 5, 6}
Properties of Probabilities
For a sample space S e e en 1 2, ,
1 0 1. ( ) for all events P e ei i
2 1 21
. ( ) ( ) ( ) ( )
= 1
P e P e P e P ei ni
n
Determine which of the following are probability models from rolling a single die.
Outcome Probability1 0.32 0.13 0.054 0.25 0.156 0.25
Not a probability model. The sum of all probabilities is not 1.
Outcome Probability1 0.22 0.23 0.24 0.25 0.26 0
All probabilities between 0 and 1 inclusive and the sum of all probabilities is 1.
Outcome Probability1 0.252 0.13 0.354 0.155 0.26 -0.05
Not a probability model. The event “roll a 6” has a negative probability.
Theorem: Probability for Equally Likely Outcomes
If an experiment has n equally likely outcomes, and if the number of ways an event E can occur is m, then the probability of E is
P EE m
n( ) Number of ways that can occur
Total number of possible outcomes
A classroom contains 20 students: 7 Freshman, 5 Sophomores, 6 Juniors, and 2 Seniors. A student is selected at random. Construct a probability model for this experiment.
P F( ) 720
P Soph( ) 520
P Jr( ) 620
P Sr( ) 220
Theorem: Additive Rule of P(E F)For any two events and
E F
P E F P E P F P E F( ) ( ) ( ) ( )
if and are mutually exclusive.
P E F P E P F
E F
( ) ( ) ( )
What is the probability of selecting an Ace or Diamond from a standard deck of cards?
P P( (Ace) =4
52 Diamond) =
1352
1
1314
P
P
(Ace or Diamond)
= (Ace) + P(Diamond) - P(Ace and Diamond)
113
14
152
1652
413
Let S denote the sample space of an experiment and let E denote an event. The complement of E, denoted E, is the set of all outcomes in the sample space S that are not outcomes in the event E.
Theorem: Computing Probabilities of Complementary Events
If E represents any event and E represents the complement of E, then
P E P E( ) ( ) 1
The probability of having 4 boys in a four child family is 0.0625. What is the probability of having at least one girl?
Sample Space: {4 boys; 3 boys, 1 girl, 2 boys, 2 girls; 1 boy, 3 girls; 4 girls}
E = “at least one girl”
E = “4 boys”
P(E) = 1 - P(E) = 1 - 0.0625 = 0.9375
What is the probability of obtaining 3 of a kind when 5 cards are drawn from a standard 52-card deck?
P(3 of a kind) CC
( , )( , )
4 3 48 4452 5
0 00325 0 325%. .This answer from the text slides is just wrong. For the correct value of 2.11% and similar examples see either of the poker sites:
http://www.math.sfu.ca/~alspach/comp18/
http://www.pvv.ntnu.no/~nsaa/poker.html
What is the probability of obtaining 3 of a kind when 5 cards are drawn from a standard 52-card deck?
P(3 of a kind))5,52(
44)2,12(413
C
C
%11.20211.0
Done correctly, one has
960,598,2
912,54
Why?