1. Probability

2. Probability experiment:
Important Terms
Probability experiment:
An action through which counts, measurements or responses are obtained
Sample space:
The set of all possible outcomes
Event:
A subset of the sample space.
Outcome:
The result of a single trial

3. Probability experiment:
Example
Probability experiment:
Roll a die
Sample space:
{ }
Event:
{ Die is even } = { }
Outcome:
{4}

4. Practice
Use a tree diagram to develop the sample space that results from rolling two six-sided dice.

5. Tree Diagrams Start 36 outcomes Two dice are rolled. Describe the
sample space.
Start
1st roll 1 2 3 4 5 6
2nd roll
36 outcomes

6. Sample Space and Probabilities
Two dice are rolled and the sum is noted.
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
Find the probability the sum is 4.
3/36 = 1/12 = 0.083
Find the probability the sum is 11.
2/36 = 1/18 = 0.056
Find the probability the sum is 4 or 11.
5/36 = 0.139

7. Ice Cream Sundaes
Consider an ice cream shop with 31 flavors of ice cream and 15 different toppings.
How many different sundaes can you make if you use 1 scoop of ice cream and two different toppings?
What if you use 3 different scoops and 2 toppings?

8. More multiplication principle
How many different outcomes can you have if you flip a coin 3 times?
How many different outcomes can you have if you flip a coin and roll a die?

9. Theoretical P(A) = number if ways A can occur total number of outcomes
In a bag you have 3 red marbles, 2 blue marbles and 7 yellow marbles. If you select one marble at random,
P(red) = 3 / (3+2+7) = 3/12 = 1/4

10. Theoretical Examples
Using a standard deck of cards, find the probability of the following.
Selecting a seven
Selecting a diamond
Selecting a diamond, heart, club or spade
Selecting a face card

11. Empirical
Probability is based on observations or experiments.

12. Empirical examples
A pond contains three types of fish: bluegills, redgills and crappies. You catch 40 fish and record the type. The following frequency distribution shows your results.
Fish Type Number of times caught
Bluegill 13
Redgill 17
Crappy 10
If you catch a fish, what is the probability that it is a redgill?

13. Empirical examples What is the probability of getting a bluegill?
What is the probability of not getting a crappy?

14. Subjective
Subjective probability results from educated guesses, intuition and estimates.
A doctor’s prediction that a patient has a 90% chance of full recovery
A business analyst predicting an employee strike being 0.25

15. Summary Classical (Theoretical) Empirical (Statistical)
The number of outcomes in a sample space is known and each outcome is equally likely to occur.
Empirical (Statistical)
The frequency of outcomes in the sample space is estimated from experimentation.
Subjective (Intuition)
Probabilities result from intuition, educated guesses, and estimates.

16. Probability If P(E) = 0, then event E is impossible.
If P(E) = 1, then event E is certain.
0  P(E)  1
Impossible Even Certain

17. Complementary Events The complement of event E is event E´.
E´ consists of all the events in the sample space that are not in event E. E E´
P(E´) = 1 - P(E)

18. Example
The day’s production consists of 12 cars, 5 of which are defective. If one car is selected at random, find the probability it is not defective.
Solution:
P(defective) = 5/12
P(not defective) = 1 - 5/12 = 7/12 = 0.583

19. Examples
What is the probability that a family with 3 children does not have 2 boys and 1 girl?
What is the probability that you do not get a pair of sixes when you roll 2 dice?

20. Probability Distributions
A discrete probability distribution lists each possible value of the random variable, together with its probability.
A survey asks a sample of families how many vehicles each owns.
number of
vehicles
Properties of a probability distribution
Each probability must be between 0 and 1, inclusive.
The sum of all probabilities is 1.

21. Example
A company tracks the number of sales new employees make each day during a 100-day probationary period. The results for one new employee are shown below. Construct and graph the probability distribution.
Sales per day, x
number of days, f

22. Example
Decide whether each distribution is a probability distribution. Explain your reasoning. x
P(x) 1/16 5/8 ¼ 3/16

23. Example
Decide whether each distribution is a probability distribution. Explain your reasoning. x
P(x)

24. Odds Odds for an event A are P(A) P(not A)
In sports we often look at wins over losses.

25. Independent Events
Two events are independent if the occurrence (or non-occurrence) of one of the events does not affect the probability of the occurrence of the other event.
Events that are not independent are dependent.

26. Examples Independent Events A = Being female B = Having type O blood
A = 1st child is a boy
B = 2nd child is a boy
Dependent Events
A = taking an aspirin each day
B = having a heart attack
A = being a female
B = being under 64” tall

27. Examples
Determine if the following events are independent or dependent.
cars are on a production line where 5 are defective and 2 cars are selected at random.
A = first car is defective
B = second car is defective.
Two dice are rolled.
A = first is a 4 and B = second is a 4

28. Multiplication Rule P(A and B) = P(A) x P(B given A) P(A) = 5/12
To find the probability that two events, A and B will occur in sequence, multiply the probability A occurs by the conditional probability B occurs, given A has occurred.
P(A and B) = P(A) x P(B given A)
Two cars are selected from a production line of 12 where 5 are defective. Find the probability both cars are defective.
A = first car is defective B = second car is defective.
P(A) = 5/12
P(B given A) = 4/11
P(A and B) = 5/12 x 4/11 = 5/33 =

29. Multiplication Rule
Two dice are rolled. Find the probability both are 4’s.
A = first die is a 4 and B = second die is a 4.

30. Independent Events
When two events A and B are independent, then P (A and B) = P(A) x P(B)
***Note for independent events P(B) and P(B given A) are the same.

31. Compare “A and B” to “A or B”
The compound event “A and B” means that A and B both occur in the same trial. Use the multiplication rule to find P(A and B).
The compound event “A or B” means either A can occur without B, B can occur without A or both A and B can occur. Use the addition rule to find P(A or B). A B A B
A or B
A and B

32. Mutually Exclusive Events
Two events, A and B, are mutually exclusive if they cannot occur in the same trial.
A = A person is under 21 years old
B = A person is running for the U.S. Senate
A = A person was born in Philadelphia
B = A person was born in Houston A
Mutually exclusive B
P(A and B) = 0
When event A occurs it excludes event B in the same trial.

33. Non-Mutually Exclusive Events
If two events can occur in the same trial, they are non-mutually exclusive.
A = A person is under 25 years old
B = A person is a lawyer
A = A person was born in Philadelphia
B = A person watches West Wing on TV
A and B
Non-mutually exclusive
P(A and B) ≠ 0 A B

34. Examples Determine whether the events are mutually exclusive or not.
Roll a die
A: Roll a 3 B: Roll a 4
Select a student
A: select a male student B: select a nursing major
Select a blood donor
A: donor is type O B: donor is female

35. Examples Select a card from a standard deck A: the card is a jack
B: the card is a face card
Select a student
A: the student is 16 years old
B: the student has blue eyes
Select a registered vehicle
A: the vehicle is a Ford
B: the vehicle is a Toyota

36. The Addition Rule
The probability that one or the other of two events will occur is: P(A) + P(B) – P(A and B)
A card is drawn from a deck. Find the probability it is a king or it is red.
A = the card is a king B = the card is red.
P(A) = 4/52 and P(B) = 26/52
but P(A and B) = 2/52
P(A or B) = 4/ /52 – 2/52
= 28/52 = 0.538

37. When events are mutually exclusive,
The Addition Rule
A card is drawn from a deck. Find the probability the card is a king or a 10.
A = the card is a king B = the card is a 10.
P(A) = 4/52 and P(B) = 4/52 and P(A and B) = 0/52
P(A or B) = 4/52 + 4/52 – 0/52 = 8/52 = 0.054
When events are mutually exclusive,
P(A or B) = P(A) + P(B)

38. Examples
1. A die is rolled. Find the probability of rolling a 6 or an odd number.
-are the events mutually exclusive?
-find P(A), P(B) and, if necessary,
P(A and B)
-use the addition rule to find the probability

39. Example
2. A card is selected from a standard deck. Find the probability that the card is a face card or a a heart.
-are the events mutually exclusive?
-find P(A), P(B) and, if necessary,
P(A and B)
-use the addition rule to find the probability

40. Contingency Table One of the responses is selected at random. Find:
The results of responses when a sample of adults in 3 cities was asked if they liked a new juice is:
Omaha
Seattle
Miami
Total
Yes
100
150
150
400 No
125
130 95
350
Undecided 75
170 5
250
Total
300
450
250
1000
One of the responses is selected at random. Find:
1. P(Miami and Yes)
2. P(Miami and Seattle)
3. P(Miami or Yes)
4. P(Miami or Seattle)

41. Contingency Table 1. P(Miami and Yes) 2. P(Miami and Seattle) = 0
Omaha
Seattle
Miami
Total
Yes
100
150
150
400 No
125
130 95
350
Undecided 75
170 5
250
Total
300
450
250
1000
One of the responses is selected at random. Find:
1. P(Miami and Yes)
2. P(Miami and Seattle)
= 250/1000 • 150/250 = 150/1000 = 0.15
= 0

42. Contingency Table 3 P(Miami or Yes) 4. P(Miami or Seattle) Omaha
Total
Yes
100
150
150
400 No
125
130 95
350
Undecided 75
170 5
250
Total
300
450
250
1000
3 P(Miami or Yes)
4. P(Miami or Seattle)
250/ /1000 – 150/1000
= 500/1000 = 0.5
250/ /1000 – 0/1000
= 700/1000 = 0.7

43. P(A and B) = P(A) • P(B given A)
Summary
For complementary events P(E') = 1 - P(E)
Subtract the probability of the event from one.
The probability both of two events occur
P(A and B) = P(A) • P(B given A)
Multiply the probability of the first event by the conditional probability the second event occurs, given the first occurred.

44. P(A or B) = P(A) + P(B) - P(A and B)
Summary
Probability at least one of two events occur
P(A or B) = P(A) + P(B) - P(A and B)
Add the simple probabilities, but to prevent double counting, don’t forget to subtract the probability of both occurring.

45. Law of Large Numbers
As an experiment is repeated over and over, the empirical probability of an event approaches the theoretical (actual) probability of the event.

46. Expected Value
Average- what you would expect given a probability distribution.
To find the expected value, multiply each event value by its probability and add

47. Example
X P(X)
0 .25
1 .50
2 .15
3 .10
0(.25) + 1(.5) + 2(.15) + 3(.10) = 1.1

48. Example
A bus arrives at a bus stop at noon, 12:20 and 1:00. You arrive at the bus stop at random times between noon and 1:00 every day so all arrival times are equally likely.
a. What is the probability that you will arrive at the bus stop between noon and 12:20? What is the mean wait time in that case?
a. 20/60 = 1/3
Wait time ranges from 0 to 20 minutes. Average wait time is 10 minutes.

49. What is your mean wait time in that case?
What is the probability that you will arrive at the bus stop between 12:20 and 1:00?
What is your mean wait time in that case?
b. 40/60 = 2/3
Wait time ranges from 0 to 40 minutes. Average wait time is 20 minutes.

50. Overall, what is your expected waiting time for the bus?
Would you expect your waiting time to be longer or shorter if the bus arrived at equally spaced intervals (say noon, 12:30 and 1:00)?
In a 30 day period you would expect 1/3 = 10 of the days to be 10 minute waits and 2/3 = 20 days to be 20 minute waits.
10 minutes x 10 days + 20 minutes x 20 days = 500 minutes
500/30 = minutes or 16 minutes 40 seconds per day.
d. If the bus arrived at equally spaced intervals the expected wait time would be 15 minutes.

51. Example
You are given 10 to 1 odds against tossing three heads in three tosses of a fair coin, meaning you win $10 if you succeed and you lose $1 if you fail.
Find the expected value (to you) of the game.
Would you expect to lose or win money on 1 game? in 100 games?

52. The probability of tossing 3 heads in 3 tosses is 1/8, and the probability of not tossing 3 heads is 7/8.
The expected value of the game is
(10 x 1/8) + (-1 x 7/8) = $0.375
-expect to win around 38 cents per game on average
The outcome of one game cannot be predicted, over 100 games you should expect to win.

53. Example
You are given 10 to 1 odds against rolling a double number (for example two 1’s or two 2’s) with the roll of two dice, meaning you win $10 is you succeed and you lose $1 if you fail.
Find the expected value (to you) of the game.
Would you expect to lose or win money on 1 game? in 100 games?
There are 6/36 chances of rolling doubles.
Probability of a double = 6/36 = 1/6
Probability of not getting a double = 5/6
(10 x 1/6) + (-1 x 5/6) = $0.83
You should expect to win 83 cents per game on average
The outcome of one game cannot be predicted, over 100 games you should expect to win.

54. Terms
The gambler’s fallacy is the mistaken belief that a streak of bad luck makes a person “due” for a streak of good luck.
The house edge is the amount the casino, or house, can expect to earn per dollar bet.

55. Example
Suppose you toss a fair coin 100 times, getting 42 heads and 58 tails, which is 16 more tails then heads.
a. Explain why, on your next toss, the difference between the number of heads and tails is as likely to grow to 17 as it is to shrink to 15.
On the 101st toss, the chances of heads is 1/2 and the chances of tails is ½ making them each equally likely. Getting a tails will raise the difference to 17, getting a tails will decrease the difference to 15.

56. How is this answer related to the gambler’s fallacy?
Extend your explanation from part a to explain why, if you toss a coin 1000 more times, the final difference is as likely to be larger than 16 as it is to be smaller than 16.
Suppose you are betting on heads with each coin toss. After the first 100 tosses, you are well on the losing side. Explain why, if you continue to bet, you will most likely remain on the losing side.
How is this answer related to the gambler’s fallacy?
b. On any subsequent toss, the chances of the tail excess growing is the same as the chances of it shrinking.
If you continue to bet, then just as seen in (b), your tail excess is as likely to be above 16 as it is to be below 16, and so your losses are as likely to exceed those already incurred b 100 tosses as they are to grow smaller.
This illustrates the gambler’s fallacy, as it shows that it is not the case that a run of bad luck must be followed by good luck.

57. Example
In a large casino, the house wins on its blackjack tables with a probability of 50.7%. All bets at blackjack are 1 to 1: If you win, you win the amount you bet, if you lose, you lose the amount you bet.
What is the expected value to you of a single game?
What is the house edge?
There are two possible outcomes in a single game, a loss for you (house wins), with a probability of 0.507, and a win for you (house loss), with a probability 1 – =
Your expected value if you bet $1 is (-1 x 0.507) + (1 x 0.493) = - $0.014
-a 1.4 cent loss. Since this is also the house’s win, per dollar gambled, the casino’s edge is 0.014

58. If you played 100 games of blackjack in an evening, betting $1 on each hand, how much would you expect to win or lose?
If you played 100 games of blackjack in an evening, betting $5 on each hand, how much would you expect to win or lose?
If you played 100 games of blackjack, then you expect to lose 100 x = $1.40
If you played 100 games of blackjack, betting $5 per game, then your loses, like your bets would increase by a factor of x 1.40 = $7.00

59. d. If patrons bet $1,000,000 on blackjack in one evening, how much should the casino expect to earn?

60. Counting and Probability7E
Counting and Probability

61. Factorial
3! = 3 x 2 x 1 = 6
5! = 5 x 4 x 3 x 2 x 1= 120
X! = x(x-1)(x-2)…1
3! Is read as “3 factorial”
On the TI-83 the factorial key is located by pressing
MATH arrow over to PROB down to 4:!

62. Arrangements
When we put things in order it is important to know if repetition is allowed.
How many 3 digit codes can be made using the numbers 0 – 9 if repetition is allowed?
If repetition is not allowed?
With repetition: 10 x 10 x 10 = 1000
Without repetition: 10 x 9 x 8 = 720

63. Examples
How many license plates can be made using 3 digits and 3 letters, repetition is allowed?
How many license plates can be made using 3 digits and 3 letters if the numbers cannot repeat?
How many 7 character passwords can be made using letters and numbers, repetition is allowed?
26 x 26 x 26 x 10 x 10 x 10 = 17,576,000
,26 x 26 x 26 x 10 x 9 x 8 = 12,654,720
36^7 = 7.84 x 10^10 = 78,364,164,096

64. The number of permutations for n objects is n!
A permutation is an ordered arrangement.
The number of permutations for n objects is n!
n! = n (n – 1) (n – 2)…..3 • 2 • 1
Emphasize that in counting permutations, order is important.
The number of permutations of n objects taken r at a time (r  n) is:

65. Example You are required to read 5 books from a list of 8.
In how many different orders can you do so?
There are 6720 permutations of 8 books reading 5.

66. Combinations
A combination is a selection of r objects from a group of n objects.
The number of combinations
of n objects taken r at a time is
In counting combinations, order does not matter. An object is either selected or it is not.

67. Example You are required to read 5 books from a list of 8.
In how many different ways can you choose the
books if order does not matter.
There are 56 combinations of 8 objects taking 5.

68. 2 4 1 3 2 1 3 1 4 1 2 3 4 3 2 4 Combinations of 4 objects choosing 2
The next two slides show the difference between permutations and combinations. With 4 objects, there are 6 ways to chose 2 without regard to order. 4 3 2 4
Each of the 6 groups represents a combination.

69. 2 4 1 3
Permutations of 4 objects choosing 2 1 2 1 2 1 3 1 3 4 4 1 1 2 2 3 3
If order is important, there are 12 ways of arranging 4 objects, choosing 2. If there are r members in each subset, then each subset can be arranged in r! ways. In this case there are 2 members in each subset so each subset (combination) can be arranged in 2! = 2*1 = 2 ways 4 4 3 3 2 2 4 4
Each of the 12 groups represents a permutation.

70. The nCr and nPr key are the Combination and Permutation keys on your calculator.
To find them on the TI-83
First type in your n, then press
MATH across to PROB
Down to nCr or nPr

71. Example
In a race with eight horses, how many ways can three of the horses finish in first, second and third place? Assume that there are no ties.
336
There are 336 possible ways that three horses can finish in first, second and third place.

72. Example
The board of directors for a company has twelve members. One member is the president, another is the vice president, another is the secretary and another is the treasurer. How many ways can these positions be assigned?
N = 12, r = 4
11,880

73. Distinguishable Permutations
The number of distinguishable permutations of n objects where n1 are one type and n2 are of another type and so on, is: n!
n1!•n2!•n3!•… nk!

74. Example To find the number of permutations in the word Mississippi
total letters = 11!
each set of repeats 2!4!4!

75. Example
A contractor wants to plant six oak trees, nine maple trees, and five poplar trees along a subdivision street. If the trees are spaced evenly apart, in how many distinguishable ways can they be planted?
N= 20, n1 = 6, n2 = 9, n3 = 5
77,597,520

76. Example
The manager of an accounting department wants to form a three-person advisory committee from the 16 employees in the department. In how many ways can the manager do this?
N = 16, r = 3
560
There are 560 different possible 3 person committees that can be selected from the 16 employees.

77. Example
A word consists of one L, two E’s, two T’s and one R. If the letters are randomly arranged in order, what is the probability the the arrangement spells the word “letter”?
1 out of 180
0.0056

78. Example
A jury consists of five men and seven women. Three are selected at random for an interview. Find the probability that all three are men.
Find the product of all the ways to choose three men from five and 0 women from 7.
Find the number of ways to choose 3 jury members from 12.
Divide a by b. 10
220
0.045