Transcript PowerPoint

Chapter 15
Counting and Probability
Section 15-1
Counting Problems and Permutations
In situations where we consider the combinations of items, or the
succession of events such as flips of a coin or the drawing of cards, each
result is called an outcome.
An event is a subset of all possible outcomes. When an event is
composed of two or more outcomes, such as choosing a card followed by
choosing another card, we have a compound event
Definition Outcome
The result of a succession of events
Definition Event
A subset of all possible outcomes.
A compound event is composed of two or more events
Theorem 15-1
The fundamental counting principle
In a compound event in which the first event can happen in n1
ways and the second event in n2 different ways and so on, and
the kth event can happen in nk different ways, the total number
of ways the composed event can happen is
n1  n2  ...  nk
Total possible number of dinners = 2(4)(2) = 16
Definition Permutation
A permutation of a set of n objects is an ordered arrangement
of the objects.
Theorem 15-2
The total number of permutations of a set of n objects is given
by
n
Pn  n  ( n  1)  ( n  2)  ...  3  2  1  n !
The set of n objects taken n at a time
Find the number of possible arrangements of the set {3,4,7}
Find the number of possible arrangements of the set {a,b,c,d}
Possible Codes:
2
2
 24
2
2
_____
_____
_____
_____
2
2
2
_____
_____
_____
 23
 30
2
2
_____
_____
 22
2
_____
 21
Eight students are to be seated in a classroom with 11 desks.
Calculate the number of seatings by choosing one of the
desks for each student.
Calculate the number of seatings by choosing one student for
each of the desks, after increasing the number of students to
11 by imagining that there are 3 “invisible” students (who are,
of course, indistinguishable).
HW #15.1
Pg 648-649 1-51 Odd
Chapter 15
Counting and Probability
15-2
Permutations For Special Counts
Objective: Find the number of permutations of n objects taken r at a time
with replacement.
Objective: Find the number of permutations of n objects taken r at a time
with replacement.
Objective: Find the number of permutations of n objects taken r at a time
with replacement.
Objective: Find permutations of a set of objects that are not all different.
How many different words (real or imaginary) can be formed
using all the letters in the word FREE?
Objective: Find permutations of a set of objects that are not all different.
How many different words (real or imaginary) can be formed
using all the letters in the word TOMORROW?
Eight students are to be seated in a classroom with 11 desks.
Calculate the number of seatings by choosing one of the
desks for each student.
Calculate the number of seatings by choosing one student for
each of the desks, after increasing the number of students to
11 by imagining that there are 3 “invisible” students (who are,
of course, indistinguishable).
Three Types of Permutations
How many ordered arrangements are there of 5 objects a, b, c, d, e
choosing 3 at a time without repetition?
How many ordered arrangements are there of 5 objects a, b, c, d, e
choosing 3 at a time with repetition?
How many ordered arrangements are there of 5 objects a, a, d, d, e
choosing 5 at a time without repetition?
If you have a group of four people that each have a different
birthday, how many possible ways could this occur?
Objective: Find circular permutations
Objective: Find circular permutations
Find the number of possibilities for each situation.
A basketball huddle of 5 players
Four different dishes on a revolving tray in the middle of a
table at a Chinese restaurant
six quarters with designs from six different states
arranged in a circle on top of your desk
HW #15.2
Pg 654-655 1-9, 11-19 Odd, 20-34
15-3
Combinations
Definition Combination
A Combination of a set of n objects is an arrangement, without
regard to order, of r objects selected from n distinct objects
without repetition, where r  n.
Find the value of each expression.
 4
a) 
 2
5
b)  
 2
n
c) 
n
n
d ) 
0 
 40 
e)  
4 
List all the combinations of the 4 colors, red, green, yellow and
blue taken 3 at a time.
How many different committees of 4 people can be formed
from a pool of 8 people?
How many ways can a committee consisting of 3 boys and 2
girls be formed if there are 7 boys and 10 girls eligible to
serve on the committee?
How many ways can a congressional committee be formed
from a set of 5 senators and 7 representatives if a committee
contains 3 senators and 4 representatives?
Winning the Lottery
In the California Mega Lottery you choose 5 different
numbers form 1 to 56 and then choose 1 number from 1 to 46
for a total of 6 numbers. How many ways can you choose
these 6 numbers?
A hamburger restaurant advertises "We Fix Hamburgers 256 Ways!“
This is accomplished using various combinations of catsup, onion,
mustard, pickle, mayonnaise, relish, tomato, and lettuce. Of course,
one can also have a plain hamburger. Use combination notation to
show the number of possible hamburgers, Do not evaluate.
HW #15.3
Pg 658 1-30
15-4 Binomial theorem
Lesson
• Pascal’s Triangle and Relation to
Combinations
• Do some expansions
• Solve for x
• Straight from the book
• HW 15.4 Pg 662-663 1-28
HW 15.4
Pg 662-663 1-28
Objective: Compute the probability of a simple event.
15-5 Probability
Definition Event
The result of an experiment is called an outcome or a simple
event. An event is a set of outcomes, that is, a subset of the
sample space.
Definition Sample Space
The set of all possible outcomes is called a sample space.
Objective: Compute the probability of a simple event.
For example
The experiment: Throwing a dart at a three-colored dart board
Sample space: Three outcomes, {red, yellow, blue}.
An Event: Hitting yellow.
When the outcomes of an experiment all have the same probability
of occurring, we say that they are equally likely.
Objective: Compute the probability of a simple event.
12 3

52 13
1 3 1 6
a ) b)  c )  1
6 6 2 6
Objective: Compute the probability of a simple event.
3 1

6 2
1
1
a) 
8! 40320
 4
 2 3
b)   
 8  14
 2
 
1 1
a) 
4 ! 24
 2
 2 1
b)   
 4 6
 2
 
Objective: Compute the probability of a simple event.
Objective: Compute the probability of a simple event.
The answers are all determined if you know which questions you will
answer correctly and which you will answer incorrectly.
Objective: Compute the probability of a simple event.
Objective: Compute the probability of a simple event.
HW #15.5
Pg 665-666 1-22
15-6
Probability of Compound Events
When you consider all the outcomes for either of two events A
and B, you form the union of A and B. When you consider only
the outcomes shared by both A and B, you form the
intersection of A and B. The union or intersection of two events
is called a Compound Event
Compound Events are considered Mutually Exclusive if the
intersection of the two events is empty.
Mutually Exclusive Events
Two Events are mutually exclusive if they cannot occur at the same
time
Example
A card is randomly selected from a standard deck of 52 cards. What
is the probability that it is an ace or a face card?
Drawing a Face Card or an Ace are mutually exclusive
Mutually Exclusive Events
Two Events are mutually exclusive if they cannot occur at the same
time
Example
A card is randomly selected from a standard deck of 52 cards. What
is the probability that it is a heart or a face card?
Drawing a Heart or a Face Card are not mutually exclusive
Objective: Find the probability that one event or another will occur.
Event A: You draw a jack or a king on a single draw from a
standard 52 card deck.
Event B: You draw a king or a diamond on a single draw from a
standard 52 card deck.
P( A)  (king  jack)
P( B)  (king  diamond)
Not
Mutually
Exclusive
Mutually
Exclusive
P( A)  P(k )  P( j )
4 4 2
P( A)   
52 52 13
P( B)  P(k)  P(d )  P(k  d )
P( B) 
4 13 1
4
 

52 52 52 13
A: Select an ace
B: Select a face card.
A and B are mutually
exclusive
SOLUTION
A: Select a heart
B: Select a Face Card
A and B are NOT
mutually exclusive
A standard six-sided number cube is rolled. Find the
probability of the given event.
1. an even number or a one
3 1 4 2
  
6 6 6 3
2. a six or a number less than 3
1 2 3 1
  
6 6 6 2
3. an even number or number greater than 5
3 1 1 3 1
   
6 6 6 6 2
4. an odd number or number divisible by 3
3 2 1 4 2
   
6 6 6 6 3
Objective: Find the probability that one event and another event will
occur.
Compute the probability of drawing a king and a queen from a wellshuffled deck of 52 cards if the first card is not replaced before the
second card is drawn.
Independent Events
The occurrence or no occurrence of one event does not effect the
probability of the other.
P(A  B) = P(A)P(B)
Dependent Events
The occurrence of the first event effects the probability of the other
event.
P(A and B) = P(A) P(B | A) = P(B) P(A | B)
By definition P(A | B) = the probability of A given that B has occurred.
Independent Events
The occurrence or no occurrence of one event does not effect the
probability of the other.
P(A  B) = P(A)P(B)
Dependent Events
The occurrence of the first event effects the probability of the other
event.
P(A  B) = P(A) P(B | A) = P(B) P(A | B)
By definition P(A | B) = the probability of A given that B has occurred.
Event A: You roll two dice. What is the probability that you get
a 5 on each die?
Event B: What is the probability you draw 2 cards from a
standard deck of 52 cards and get two aces?
P( A)  (5  5)
P( B)  (Ace  Ace)
Independent
P( A)  P(5)  P(5)
1 1 1
P( A)   
6 6 36
Dependent
P( B)  P( Ace) P(Aceon 2nd |Aceon 1st )
P( B) 
4 3
1
 
52 51 221
Event B: What is the probability you draw 2 cards from a
standard deck of 52 cards and get two aces?
P( B)  (Ace  Ace)
Dependent
Sample Space Method:
#of ways todraw twoaces
P( B) 
#of ways todraw twocards
P( B)  (Aceon 2nd card|Aceon 1st card)
4 3
1
P( B)   
52 51 221
 4
 2
1



P( B) 
221
 52 
2 
 
Compute the probability of drawing a king and a queen from a wellshuffled deck of 52 cards if the first card is not replaced before the
second card is drawn.
Multiplication Rule Method:
P(king on 1st  queen on 2nd) =
= P(king on 1st)  P(queen on 2nd, given king on 1st)
4 4
4
 

52 51 663
Sample Space Method:
P  4,1  P  4,1
4

P(king on 1st card  queen on 2nd card) =
P  52, 2 
663
Multiplication rule order matters
Probability of a king and
queen in that order
SOLUTION
A: Getting more than $500 on the first spin
B: Going bankrupt on the second spin.
The two events are independent.
Find the probability of the given events.
1. An 8 on the first roll and doubles on the second roll?
P (8)  P ( Doubles ) 
5 6
5
 
36 36 216
2. An even sum on the first roll and a sum greater than 8
on the second roll.
18 10 5
P ( Even)  P ( Sum  8)   
36 36 36
3. Drawing a 5 on the first draw and a king on the second
draw without replacement.
4 4
16
4
P (5)  P ( K | 5)   

52 51 2652 663
HW #15.6a
Pg 670-671 1-23 Odd
15-6 Day 2
Probability of Compound Events
PROVE: If A and B are independent, then P(B | A) = P(B)
What is the probability that in a room of 40 people 2 or more people
have the same birthday?
Objective: Compute the probability of a simple event.
Rolling Dice Two six-sided dice are rolled. Find the probability of the
given event.
The sum is even and a multiple of 3.
P( Even  mult 3)  P(even) P(mult 3 | even)
18 6
6 1



36 18 36 6
The sum is not 2 or not 12.
Objective: Compute the probability of a simple event.
Rolling Dice Two six-sided dice are rolled. Find the probability of the
given event.
The sum is greater than 7 or odd.
The sum is prime and even.
Objective: Compute the probability of a simple event.
A standard deck of cards contains 4 suits (heart, diamond, club,
spade) and 13 cards per suit (Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack,
Queen, King). Suppose five cards are drawn from the deck
without replacement. What is the probability that the cards will be
the 10, jack, queen, king, and ace of the same suit?
 4  5
1   5 
   
 52 
5 
 
If P  A  0.28, P  B   0.41 , and P  A  B   0.16 , what is P  A  B  ?
HW #15.6b
Pg 670-671 2-22 Even, 24, 26-29
More Fun With Probability
Objective: Compute the probability of a simple event.
Suppose 5 cards are drawn from a deck of 52
cards. What is the probability that you draw 5
Spades?
13 
5 
   1287
 52  2598960
5 
 
Objective: Compute the probability of a simple event.
Suppose 5 cards are drawn from a deck of 52
cards. What is the probability that you draw 2
cards with the same number and 3 other
different cards?
 13   4  12   4 
1   2   3   
      1 
 52 
5 
 
 4  4 
1  1 
  
Objective: Compute the probability of a simple event.
Suppose 5 cards are drawn from a deck of 52
cards. What is the probability that you draw 3
cards with the same number and 2 other
different cards?
 13   4  12   4 
1   3   2   
      1 
 52 
5 
 
 4
1 
 
Objective: Compute the probability of a simple event.
Objective: Compute the probability of a simple event.
HW#R-15
Pg 685-686 1-22
Definition: Expected Value
• In probability theory the expected value of a random
variable is the sum of the probability of each possible
outcome of the experiment multiplied by its value.
– Represents the average amount one "expects" to win per
game if bets with identical odds are repeated many times.
– Note that the value itself may not be expected in the general
sense; It may be unlikely or even impossible.
– A game or situation in which the expected value for the
player is zero (no net gain nor loss) is called a “fair game”.
Example
Find the expected value of X, where the values of X and
their corresponding probabilities are given by the following
table: xi
2
5
9
24
pi
0.4
0.2
0.3
0.1
SOLUTION
E ( X )  0.4  2  0.2  5  0.3  9  0.1 24
 0.8  1.0  2.7  2.4
 6.9
Ten cards of a children’s game are numbered with all
possible pairs of two different numbers from the set
{1, 2, 3, 4, 5}. A child draws a card, and the random
variable is the score of the card drawn. The score is 5 if
the two numbers add to 5; otherwise, the score is the
smaller number on the card. Find the expected score of a
card.
Two numbers can be selected from the
five numbers in C(5, 2) = 10 ways
Score
5
1
2
3
4
Pairs
(1, 4), (2, 3)
(1, 2),(1, 3), (1, 5)
(2, 4), (2, 5)
(3, 4), (3, 5)
(4, 5)
E(X) = 1(3/10) + 2(2/10) + 3(2/10) + 4(1/10) + 5(2/10) = 2.7
Expected Value Roulette
For example, an American roulette wheel has 38 equally possible
outcomes. A bet placed on a single number pays 35-to-1 (this means
that you are paid 35 times your bet and your bet is returned, so you get
36 times your bet). So the expected value of the profit resulting from a
$1 bet on a single number is, considering all 38 possible
outcomes:
which is about −$0.0526. Therefore one expects, on average, to lose
over five cents for every dollar bet.
Expected Value
Suppose you work for an insurance company, and you sell a
$10,000 whole-life insurance policy at an annual premium of
$290. Actuarial tables show that the probability of death during the
next year for a person of your customer's age, sex, health, etc., is
.001. What is the expected gain (amount of money made by the
company) for a policy of this type?
Gain x
Sample Point Probability
$290
Customer lives
$290-$10,000 Customer dies
.999
.001
If the customer lives, the company gains the $290 premium as
profit. If the customer dies, the gain is negative because the
company must pay $10,000, for a net "gain" of $(290 - 10,000).
Expected Value
Gain x
Sample Point Probability
$290
Customer lives
$290-$10,000 Customer dies
.999
.001
E   xP ( x)  290(.999)  (290  10000)(.001)
 280
The insurance company expects to make a $280 profit on the deal
at the end of the first year.
Expected Value
The chance of winning Florida’s pick six lottery is about 1 in
14,000,000. Suppose you buy a $1.00 lotto ticket in anticipation
of the $7,000,000 jackpot. Calculate your expected net winnings.
Gain x
Sample Point
Probability
7,000,000
-1
Win
Lose
1/14,000,000
13999999/14000000
1

 13,999,999 
E   xP ( x)  7,000,000 
 (0) 


 14,000,000 
 14,000,000 
 $0.50
The college hiking club is having a fundraiser to buy a new
toboggan for winter outings. They are selling Chinese fortune
cookies for 35 cents each. Each cookie contains a piece of paper
with a different number written on it. A random drawing will
determine which number is the winner of a dinner for two at a local
Chinese restaurant. The dinner is valued at $40. Since the fortune
cookies were donated to the club, we can ignore the cost of the
cookies. The club sold 816 cookies. John bought 12 cookies.
1. What is the probability he will win
2. What is the probability he will loose?
3. What is the expected value of the game?
Binomial Probability Model
On a TV quiz show each contestant has a try at the
wheel of fortune. The wheel of fortune is a roulette
wheel with 36 slots, one of which is gold. If the
ball lands in the gold slot, the contestant wins
$50,000. No other slot pays. What is the
probability that the quiz show will have to pay the
fortune to three contestants out of 100?
Binomial Probability
• There are only two outcomes – Success or Failure
• There is a number of fixed trials
• All trials are independent and repeated under
identical conditions
• For each trial the probability of success is the
same and P(success) + P(Failure) = 1
• Looking for the P(r successes out of n trials)
n
r
P(r succesesin n trials)     P( S )  ( P( F )) nr
r 
Wheel of Fortune Problem
• Each of the 100 contestants has a trial so
there are 100 trails (n = 100)
• The trials are independent (assuming a fair
wheel)
• Only two outcomes win or lose
• P(Success) = 1/36
• P(Failure) = 35/36
• P (Success) + P(Failure) = 1
3
1003
100

  1   35 
P(3succesesin100 trials)  
  36   36 
3

   
A fair quarter is flipped three times. Find the following probabilities:
1. You get exactly three heads  3   1   1  1
 3  2   2   8
    
3
2. You get exactly two heads
0
3  1   1  3
 2 2   2   8
    
2
1
3. You get two or more heads  3   1 2  1 1  3   1 3  1 0 1
 2   2   2    3  2   2   2
         
4. You get exactly three tails
3  1   1  1
 0 2   2   8
    
0
3
Joe Blow has just been given a 10 question multiple choice quiz
in history class. Each question has 5 answers of which one is
correct. Since Joe has not attended class recently, he does not
know any of the answers. Assuming Joe guesses on all 10
questions, find the indicated probabilities
10
0
10
  1   4 
1

1. Joe gets all 10 correct      
10   5   5  9765625
10   1   4 
410
2. Joe gets all 10 wrong        10
5
 0  5   5 
0
10
10   1   4 
5  5   5 
    
5
3. Joe gets at half correct
4. Joe gets at least one correct
5
0
10
10
  1   4 
1     
 0  5   5 
Binomial expected value handout