#### Transcript Document

```Chapter 9
Counting, Probability
Distribution and Further
Topics on Probability
9.1 Probability Distributions and
Expected Value
A random variable is a function that assigns a real
number to each outcome of an experiment.
Example: “number of heads” showing when 2 coins
are tossed
• Probability distributions
Table with all possible values of random
variables, together with the corresponding
probabilities.
• Histogram: bar graph used to display probability
distributions.
Histogram
X
1
2
3
4
5
P(x)
.13
.29
.38
.13
.08
0.38
0.40
0.35
0.29
0.30
0.25
0.20
0.15
0.13
0.13
0.08
0.10
0.05
0.00
1
2
3
4
5
P(x  4) = P(x=4) + P(x=5) = .13 + .08 = .21
Examples
1) Give the probability distribution for the
number of heads showing when 2 coins
are tossed, draw the histogram, and find
the probability that at least one coin
Examples
2) Give the probability distribution for the
following experiment: 2 dice are rolled,
the total number of points are recorded.
Examples
3) Give the probability distribution for the
following experiment: 3 cards are drawn
from a deck, and number of faces is
recorded.
EXPECTED VALUE
Suppose the random variable x can take
on the n values x1, x2, …, xn.
Also, suppose the probabilities that these
values occur are respectively p1, p2, …, pn.
Then the expected value of the random
variable is:
E(x) = x1p1 + x2p2 + … + xnpn
Example
Find the expected value of x in the
probability distribution below:
X
P(x)
1
.13
2
.29
3
.38
4
.13
5
.08
Examples
Example 3 (p. 513)
A local church decides to raise money by
raffling off a micro oven worth \$400, a dinner
for two worth \$80, and 2 books worth \$20. A
total of 2000 tickets are sold at \$1 each.
Find the expected value of winnings for a
person who buys 1 ticket in the raffle.
Examples
Example 4 (p. 514)
Hoa and Nam toss a coin to see who buys
the coffe (\$1.25 a cup). Hoa tosses and
Nam calls the outcome. If Nam is correct,
Hoa buys the coffee; otherwise Nam pays.
Find expected winnings of Hoa.
Examples
Example 5 (p.515)
Bob is 50 years old. He must decide 1 of 2
options from an insurance company: get
\$60000 at age 60 or \$65000 at age 70.
Given that probability of person living from
age 50 to 60 is .88, and probability of person
living from age 50 to 70 is .64. Use expected
value to help Bob to select which option is
more valuable.
Examples
Example 7 (p. 516): Suppose a family has 3
children.
a) Find the probability distribution for the
number of girls
b) Use the distribution from part a) to find
the expected number of girls in a 3-child
family.
9.2 The Multiplication Principle,
Permutations and Combinations
MULTIPLICATION PRINCIPLE
• Suppose n choices must be made, with
m1 ways to make choice 1,
and for each of these,
m2 ways to make choice 2,
and so on, with
mn ways to make choice n.
• Then there are m1m2…mn
different ways to make the entire sequence of
choices.
Examples
Example 1 (p. 521) Angela has 9 skirts, 8
blouses and 13 different pairs of shoes. How
many different skirt –blouse-shoe choices
does she have?
Example 3 (p. 522) A combination lock can
be set to open to any 3-letter sequence.
a) How many sequences are possible?
b) How many sequences are possible if no letter
is repeated?
n-FACTORIAL
For any natural number n, we define:
n! = n(n-1)(n-2)…(3)(2)(1).
0! is defined to be the number 1.
Example:
10!
1) Evaluate
9!
5!
2) Evaluate
2!3!
Examples
• Example 8 (p. 524). A teacher has 5 books
and wants to display 3 of them side by
side on her desk. How many
arrangements of 3 books are possible?
PERMUTATION
Definition:
• A permutation of n elements is an ordering of the
elements.
The number of permutations of an n-element set is n!
• A permutation of r elements (r1) from a set of n
elements is an ordering of the r elements.
Number Of Permutations P(n,r):
• If P(n, r) , r  n, is the number of permutations of
n elements taken r at a time, then
n!
P(n, r) =
(n  r )!
Examples
• Example 9: In 2004, 8 candidates sought for the
Democratic nomination for president. In how
many ways could voters rank their first, second
and third choice?
• Example 10: A televised talk show will include 4
women and 3 men as panelists.
a)In how many ways can the panelists be seated
in a row of 7 chairs?
b) In how many ways can the panelists be seated
if the men and women are to be alternated?
c) In how many ways can the panelists be seated if
men must sit together and the women must sit
together?
COMBINATION
Definition:
• A subset of items selected without
regard to order is called combination.
Number Of Combinations:
• If C(n, r) , r  n, is the number of
combinations of n elements taken r at a
n!
time, then C(n, r) =
r!(n  r )!
Examples
Example 11 (p. 527):
From a group of 10 students, a committee is
to be chosen to meet with the dean. How
many different 3-person committee are
possible?
Example 12 (p. 527)
In how many ways can a 12-person jury be
chosen from a pool of 40 people?
• Example 13
Examples
Example 13 (p. 527)
Three managers are to be selected from a group
of 30 to work on a special project.
(a) In how many different ways can the managers
se selected?
(b) In how many different ways can the group of 3
be selected if a certain manager must work on
the project?
(c) In how many different ways can a nonempty
group of at most 3 managers be selected from
these 30 managers?
Examples
Example 14 (p. 528): permutation or combination?
(a) How many 4-digit code numbers are possible if
no digits are repeated?
(b) A sample of 3 lightbulbs is randomly selected
from a batch of 15 bulbs. How many different
samples are possible?
(c) In a basketball conference with 8 teams, how
many games must be played so that each team
plays every other team exactly once?
(d) In how many ways can 4 patients be assigned
to 6 hospital rooms so that each patient has a
private room?
Examples
Example 15 (p. 529)
A manager must select 4 employees for
promotion. 12 are eligible.
(a) In how many ways can the 4 be chosen?
(b) In how many ways can 4 employees be
chosen to be placed in 4 different jobs?
Example 17 (p. 530)
Five cards are dealt from a standard 52-card deck.
(a) How many hands have all face cards?
(b) How many hands have have exactly 2 hearts?
(c) How many 5-card hands have all cards of a
single suit?
9.3 APPLICATIONS OF COUNTING
Example 1 (p. 535)
From a box containing 1
red, 3 white and 2 green
marbles, two marbles are
drawn one at a time without
replacing the first before
the second is drawn. Find
the probability that one
white and one green
marble are drawn.
9.3 APPLICATIONS OF COUNTING
Example 2 (p. 535).
1) How many ways can we select 4
students in a class of 15 students?
2) One student in the class is Nam. Find
the probability that Nam will be among
the 4 selected students.
9.3 APPLICATIONS OF COUNTING
Example 3 (p. 536).
A container has 12 items, in which there are 2
defective. We test 3 items at random; if any of the
3 is defective, the container will not be shipped
until each item is checked. Find the probability
that this container will not be shipped.
9.3 APPLICATIONS OF COUNTING
Example 4 (p. 537).
5 cards are drawn from a
deck. Find:
1) P(heart flush)
2) P(any flush)
3) P(full house of aces and
kings)
4) P(any full house)
9.3 APPLICATIONS OF COUNTING
• Example 5 (p. 538)
A cooler contains 8 different kinds of soda,
among which 3 cans are Pepsi, Classic coke
and Sprite. What is the probability, when
picking at random, of selecting the 3 cans in the
particular order listed in the previous sentence?
• Example 6 (p. 538)
There are 5 people in a room. Find the
probability that at least 2 of the people have the
same birthday.
9.4 BINOMIAL PROBABILITY
BINOMIAL EXPERIMENT
• The same experiment is repeated A a
fixed number of times.
• There are only two possible outcomes,
success or failure.
• The probability of success for each trial is
constant.
• The repeated trials are independent.
Example
• Find the probability of getting 5 ones on 5
rolls of a die.
• Find the probability of getting 4 ones on 5
rolls of a die.
• Find the probability of getting 3 ones on 5
rolls of a die.
BINOMIAL PROBABILITY
If p is the probability of success in a single
trial of a binomial experiment, the
probability of x successes and n–x failure
in n independent repeated trials of the
experiment is:
C(n, x) px (1 – p)n–x
Example 1 – 2 (p. 544 –545)
Expected Value for
Binomial Probability
When an experiment meets the four
conditions of a binomial experimenet with
n fixed trials and constant probability of
success p, the expected value is:
E(x) = np
Example 3 – 5 (p. 545 –546)
9. 4 MARKOV CHAINS
MARKOV CHAIN
• One kind of stochastic process, in
which the outcome of an experiment
depends only on the outcome of the
previous experiment.
• Example 1 (p. 549)
• Transition diagram
• Transition matrix
Transition diagram
0.2
0.8
A
40%
0.65
B
60%
A
0.35
B
A  0.8 0.2 
B 0.35 0.65
Transition matrix
A transition matrix has the following
features:
1. It is square, since all possible states
must be used both as rows and as A
columns.
2. All entries are between 0 and 1, A  0.8
inclusive, because all entries

B 0.35
represent probabilities.
3. The sum of the entries in any row
must be 1, because the numbers in
the row give the probability of
changing from the state at the left to
one of the states indicated across
the top.
B
0.2 

0.65
Regular Transition Matrix
A transition matrix is regular if some power of the
matrix contains all positive numbers.
Example:
.75 .25 0 
A   0 .5 .5 is regular
 .6 .4 0 
.5 0 .5
B   0 1 0  is not regular
 0 0 1 
A Markov chain is a regular Markov chain if its
transition matrix is regular
Probability Vector
• Example 2 (p. 476)
• Probability vector: matrix of only one row,
having nonnegative entries with the sum of
the entries equal to 1.
• Example of probability vectors:
– [0.40, 0.60]
– [0.53, 0.47]
– [0.40, 0.25, 0.35]
Equilibrium Vector
Suppose a regular Markov chain has a
transition matrix P.
1. For any initial probability vector v, the products
vPn approach a unique vector V as n gets
larger and larger. Vector V is called the
equilibrium or fixed vector.
2. Vector V has the property that VP = V.
3. Vector V is found by solving a system of
equations obtained from the matrix equation
VP = V and from the fact that the sum of the
entries of V is 1.
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