1. Chapter 5 Probability in Our Daily Lives
Section 5.1
How Probability Quantifies Randomness

2. Probability—outline:
I Introduction to Probability
A Satisfactory outcomes vs. total outcomes
B Basic Properties
C Terminology
II Combinatory Probability
A The Addition Rule – “Or”
The special addition rule (mutually exclusive events)
The general addition rule (non-mutually exclusive events)
B The Multiplication Rule – “And”
The special multiplication rule (for independent events)
The general multiplication rule (for non-independent events)
C The Complement Rule – “Not”

3. Random Phenomena For random phenomena, the outcome is uncertain.
In the short-run, the proportion of times that something happens is highly random.
In the long-run, the proportion of times that something happens becomes very predictable.
As we make more observations, the proportion of times that a particular outcome occurs gets closer and closer to a certain number we would expect.
Probability quantifies long-run randomness.

4. Independent Trials
Different trials of a random phenomenon are independent if the outcome of any one trial is not affected by the outcome of any other trial.
Example:
If you have 20 flips of a coin in a row that are “heads”, you are not “due” a “tail” - the probability of a tail on your next flip is still 1/2. The trial of flipping a coin is independent of previous flips.

5. Randomness and probability
A phenomenon is random if individual outcomes are uncertain, but there is nonetheless a regular distribution of outcomes in a large number of repetitions.
Examples: Tossing a coin, rolling a die, choosing a card
The probability of any outcome of a random phenomenon can be defined as the proportion of times the outcome would occur in a very long series of repetitions – this is known as the relative frequency definition of probability.

6. Examples of Randomness and probability
Chance behavior is unpredictable in the short run but has a regular and
predictable pattern in the long run.
Random: individual outcomes are uncertain but there is a regular distribution of outcomes in a large number of repetitions
Relative frequency (proportion of occurrences) of an outcome settles down to one value over the long run. That one value is then defined to be the probability of that outcome.
1 Reason to study probability:
(1) It serves as the foundation of Statistics. (For expectation, Confidence interval and Hypothesis Testing, they all need a good understanding of it.)
(2) It helps us to understand random mechanism.
2.Some interesting questions:
(1) Suppose you tossed a fair coin 10 times and surprisingly all you’ve got were heads. If you want to make another toss, will it more likely end up with a tail?
(2) With a full deck of well shuffled cards (52 cards) facing down, if you put the top card aside without turning over and looking at it, what’s the probability that the next card that you deal is a king?

7. Law of Large Numbers
As the number of trials increase, the proportion of occurrences of any given outcome approaches a particular number “in the long run”.
For example, as one tosses a die, in the long run 1/6 of the observations will be a 6.
We will interpret the probability of an outcome to represent long-run results.

8. Coin toss: Law of Large Numbers
The result of any single coin toss is random. But the result over many tosses is predictable, as long as the trials are independent (i.e., the outcome of a new coin flip is not influenced by the result of the previous flip).
Coin toss: Law of Large Numbers
The probability of heads is 0.5 = the proportion of times you get heads in many repeated trials.
First series of tosses
Second series

9. Probability
With random phenomena, the probability of a particular outcome is the proportion of times that the outcome would occur in a long run of observations.
Example:
When rolling a die, the outcome of “6” has probability = 1/6. In other words, the proportion of times that a 6 would occur in a long run of observations is 1/6.

10. Probability models
Probability models describe mathematically the outcomes of random processes and consist of two parts:
1) S = Sample Space: This is a set, or list, of all possible outcomes of the random process. An event is a subset of the sample space. 2) A probability assigned for each possible event in the sample space S – the assignment should make sense in a model meant to describe the real world ...
Example: Probability Model for a Fair Die:
S = {1, 2, 3, 4, 5, 6}
Probability of odd # = 0.5
Probability of even # = 0.5

11. Sample space Find the sample space for 1. Tossing two coins
2. Rolling two dice
This is S:
{(1,1), (1,2), (1,3), ……etc.}
There are 36 possible outcomes in S, all equally likely (given fair dice).

12. Probability Notation An event is denoted by A, B, C…
Eg: A=getting a heart from a full deck of cards.
B=getting a head when flipping a coin.
If A is an event, then P(A) stands for the probability that event A occurs. It is read “the probability of A”.
Probability for Equally Likely Outcomes
Suppose an experiment has N possible outcomes, all equally likely. Probability that event A occurs equals:

13. Use relative frequency to find Probability (sample point)
Eg1: Find the following probabilities:
Pr(Red)
Pr(Blue)
Pr(Green)
Pr(Yellow)
Eg2: Roll a dice labeled 1 to 6. Which of the following is LEAST LIKELY?
an even number?
an odd number?
a number greater than 4?
a number less than 4?

14. Probability rules
1) Probabilities range from 0 (no chance of the event) to 1 (the event has to happen).
For any event A, 0 ≤ P(A) ≤ 1
Probability of getting a Head = 0.5
We write this as: P(Head) = 0.5
P(neither Head nor Tail) = 0
P(getting either a Head or a Tail) = 1
2) Because some outcome must occur on every trial, the sum of the probabilities for all possible outcomes (the sample space) must be exactly 1.
P(sample space) = 1
Coin toss: S = {Head, Tail}
P(head) + P(tail) = =1
 P(sample space) = 1
Property 1: The probability of an event that cannot occur is 0. (An event that cannot occur is called an impossible event.)
Property 2: The probability of an event that must occur is 1. (An event that must occur is called a certain event.)

15. A deck of playing cards Four Suits: Spade , Heart, Club, Diamond;
13 Ranks: Ace, Deuce, 3 ,4, 5, 6, 7, 8, 9, 10, Jack, Queen, King.

16. Eg1: Event A = the king of hearts is selected
P(A)=1/52
Eg2: Event B = a king is selected
P(B) = 4/52=1/13

17. Eg3: Event C = a heart is selected
P(C) = 13/52=1/4
Eg4: Event D = a face card is selected
P(D) =12/52=3/13

18. Relationships Among Events
Venn diagram for event E:
(not E): The event that “E does not occur.”
(A & B): The event that “both A and B occur.”
(A or B): The event that “either A or B or both occur”  “at least one of two occur”.

19. Probability rules (cont. )
Venn diagrams:
A and B disjoint
3) Addition rule for disjoint events:
If events A and B are disjoint, then
P(A or B) = “P(A U B)” = P(A) + P(B)
Note: Two events A and B are disjoint or mutually exclusive if they have no outcomes in common and can never happen together. The probability that A or B occurs is then the sum of their individual probabilities.
A and B not disjoint
Example: If you flip two coins, and the first flip does not affect the second flip: S = {HH, HT, TH, TT}. The probability of each of these events is 1/4, or 0.25.
The probability that you obtain “only heads or only tails” is: P(HH or TT) = P(HH) + P(TT) = = 0.50

20. Probability rules (cont.)
Coin Toss Example:
S = {Head, Tail}
Probability of heads = 0.5
Probability of tails = 0.5
Probability rules (cont.)
4) Multiplication rule for independent events:
If events A and B are independent, then P(A and B) = P(A)*P(B)
Note: Two events A and B are independent if knowing that one occurs does not change the probability that the other occurs.
Two consecutive coin tosses:
P(first Tail and second Tail) = P(first Tail) * P(second Tail) = 0.5 * 0.5 = 0.25
Venn diagram:
Event A and event B. The intersection represents the event {A and B} and outcomes common to both A and B.

21. Probability rules (cont.)
Coin Toss Example:
S = {Head, Tail}
Probability of heads = 0.5
Probability of tails = 0.5
Probability rules (cont.)
5) Complement rule : the probability of an event not occurring is 1 minus the probability that is does occur.
P(not A)=1-P(A)
The complement of any event A is the event that A does not occur, written as Ac.
That is: P(not A) = P(Ac) = 1 − P(A)
Tailc = not Tail = Head
P(Tailc) = 1 − P(Tail) = 0.5
Venn diagram:
Sample space made up of an event A and its complementary Ac, i.e., everything that is not A.

22. M&M candies
If you draw an M&M candy at random from a bag, the candy will have one of six colors. The probability of drawing each color depends on the proportions manufactured, as described here:
Q1: What is the probability that an M&M chosen at random is blue?
Color
Brown
Red
Yellow
Green
Orange
Blue
Probability
0.3
0.2
0.1 ?
Q2: What is probability that a random M&M is any of red, yellow, or brown?
Q3: What is probability that a random M&M is neither red, yellow, nor brown?
Q4: If two cadies are drawn with replacement, what is probability that the first
M&M is brown, and the second M&M is yellow?

23. M&M candies
If you draw an M&M candy at random from a bag, the candy will have one of six colors. The probability of drawing each color depends on the proportions manufactured, as described here:
What is the probability that an M&M chosen at random is blue?
Color
Brown
Red
Yellow
Green
Orange
Blue
Probability
0.3
0.2
0.1 ?
S = {brown, red, yellow, green, orange, blue}
P(S) = P(brown) + P(red) + P(yellow) + P(green) + P(orange) + P(blue) = 1
P(blue) = 1 – [P(brown) + P(red) + P(yellow) + P(green) + P(orange)]
= 1 – [ ] = 0.1
What is the probability that a random M&M is any of red, yellow, or brown?
P(red or yellow or orange) = P(red) + P(yellow) + P(brown)
= = 0.7
What is the probability that a random M&M is neither red, yellow, nor brown?

24. Chapter 5 Probability in Our Daily Lives
Section 5.2
Finding Probabilities – A deeper look

25. Sample Space
For a random phenomenon, the sample space is the set of all possible outcomes.
Figure 5.3 Tree Diagram for Student Performance on a Three-Question Pop Quiz. Each path from the first set of two branches to the third set of eight branches determines an outcome in the sample space. Question: How many possible outcomes would there be if the quiz had four questions?
From the tree diagram, a student’s performance has eight possible outcomes: {CCC, CCI, CIC, CII, ICC, ICI, IIC, III}

26. Event
An event is a subset of the sample space. An event corresponds to a particular outcome or a group of possible outcomes.
Example:
For a student taking the three-question pop quiz, some possible events are:
Event A = a student answers all 3 questions correctly = (CCC)
Event B = a student passes (at least 2 correct) = (CCI, CIC, ICC, CCC)

27. Dice You toss two dice. This is S: Event A=sum is 5
What is the sample space?
What is the event that the sum of the numbers is 5?
This is S:
{(1,1), (1,2), (1,3), ……etc.}
There are 36 possible outcomes in S, all equally likely (given fair dice). Thus, the probability of any one of them is 1/36.
Event A=sum is 5
={(4&1),(3&2),(2&3),(1&4)}

28. SUMMARY: Probability of an Event Rule #1
The probability of an event A, denoted by P(A), is obtained by adding the probabilities of the individual outcomes in the event.
When all the possible outcomes are equally likely,

29. Review: Contingency table Example: Tax Audit
Table 5.2 Contingency Table Cross-Tabulating Tax Forms by Income Level and Whether Audited or Not. There were million returns filed. The frequencies in the table are reported in thousands. For example, 1260 represents 1,260,000 tax forms that reported income under $200,000 and were audited.

30. Example: Tax Audit
1. What is the sample space for selecting a taxpayer?
{(under $200,000, Yes), (under $200,000, No), ($200, $1,000,000, Yes),...,6. (More than $1,000,000, No)}
2. For a randomly selected taxpayer in 2002:
What is the probability of an audit?
1413/138,242 =
What is the probability of an income of more than $1,000,000?
393/138,242 =

31. Basic rules for finding probabilities about a pair of events
Some events are expressed as the outcomes that:
are not in some other event (complement of the event), which is P(not A).
are in one event and in another event (intersection of two events), which is P(A and B).
are in one event or in another event or in both events (union of two events), which is P(A or B).

32. Intersection of Two Events
The intersection of A and B consists of outcomes that are in both A and B.
Figure 5.6 The Intersection and the Union of Two Events. Intersection means A occurs and B occurs, denoted “A and B.” The intersection consists of the shaded “overlap” part in Figure 5.6 (a). Union means A occurs or B occurs or both occur, denoted “A or B.” It consists of all the shaded parts in Figure 5.6 (b). Question: How could you find P(A or B) if you know P(A), P(B), and P(A and B)?

33. Union of Two Events
The union of A and B consists of outcomes that are in A or in B or in both A and B.
Figure 5.6 The Intersection and the Union of Two Events. Intersection means A occurs and B occurs, denoted “A and B.” The intersection consists of the shaded “overlap” part in Figure 5.6 (a). Union means A occurs or B occurs or both occur, denoted “A or B.” It consists of all the shaded parts in Figure 5.6 (b). Question: How could you find P(A or B) if you know P(A), P(B), and P(A and B)?

34. Probability of the Union of Two Events Rule #3
Addition Rule:
For the union of two events,
P(A or B) = P(A) + P(B) – P(A and B)
If the events are disjoint(mutually exclusive),
P(A and B) = 0, so P(A or B) = P(A) + P(B)
Figure 5.7 The Probability of the Union, Outcomes in A or B or Both. Add P(A) to P(B) and subtract P(A and B) to adjust for outcomes counted twice.

35. Example: Probability of the Union of Two Events
Consider a family with two children. The sample space possibilities for the genders of the two children are {FF, FM, MF, MM}. Assume the four outcomes in the sample space are equally likely.
Let A = {first child a girl} and B = {second child a girl}
Then
P(A) = P({FF, FM}) = 0.50,
P(B) = P({FF, MF}) = 0.50, and
P(A and B) = P({FF}) = 0.25 →This indicates that A and B are not disjoint!
The event A or B is the event that the first child is a girl, or the second child is a girl, or both are girls.
In other words, that at least one child is a girl. Its probability is:
P(A or B) = P(A) + P(B) - P(A and B) = = 0.75.

36. Example: The Disjointness
1. Consider a family with two children. The sample space possibilities for the genders of the two children are {FF, FM, MF, MM}. Assume the four outcomes in the sample space are equally likely.
Let A = {first child a girl} and C = {first child a boy}
Then event A and C are disjoint,
since P(A and C)=P(first child is both a girl and a boy)=0.
2. For a student taking the three-question pop quiz with each question has one out of five correct answers.
Let A = {1st ans is correct } and B = {1st answer is incorrect}
Then event A and B are disjoint,
since P(A and B)=P(1st ans is both correct and incorrect)=0.
Let C = {2nd answer is incorrect}
Then event A and C are NOT disjoint,
since P(A and C)=P(1st ans is correct and 2nd answer is correct)≠0.

37. Probability of the Intersection of Two Independent Events. Rule #4.
Multiplication Rule:
For the intersection of two independent events, A and B, P(A and B) = P(A) x P(B)
For instance, for two rolls of a die,
P(6 on roll 1 and 6 on roll 2) = P(6 on roll 1) x P(6 on roll 2) = 1/6 x 1/6 = 1/36
For instance, for the gender of two children,
P(1st is a Girl and 2nd is a Girl)= P(1st is a Girl) x P(2nd is a Girl)=1/2 x1/2=1/4
This multiplication rule extends to more than two independent events.

38. Example: Guessing yet Passing a Pop Quiz
What is the probability of getting 3 questions correct by guessing? If each question has five options, then the probability of selecting the correct answer for any given question is 1/5, or 0.20.
The probability that the student answers all three questions correctly is:
P(C&C&C) = P(C) * P(C) * P(C) = 0.20 * 0.20 * 0.20 =

39. Example: Guessing yet Passing a Pop Quiz
Figure 5.8 is a tree diagram showing how to multiply probabilities to find the probabilities for all eight possible outcomes.
Figure 5.8 Tree Diagram for Guessing on a Three-Question Pop Quiz. Each path from the first set of branches to the third set determines one sample space outcome. Multiplication of the probabilities along that path gives its probability, when trials are independent.

40. Example: Guessing yet Passing a Pop Quiz
Q: What is the probability that a student answers at least 2 questions correctly?
Ans:
P(CCC) + P(CCI) + P(CIC) +P(ICC) = (0.032) =0.104
In summary, there is only about a 10% chance of passing, when a student randomly guesses the answers.

41. Events Often Are Not Independent
In practice, make sure that assuming independence is realistic.
Don’t assume that events are independent unless you have given this assumption careful thought and it seems plausible.

42. Example: Events Often Are Not Independent
A survey asked 100 respondents whether they favored or opposed a new gun law. The responses, categorized by gender, are given in the table below:
sex
New gun law M F
Total
Favor 30 60
oppose 20 40 50
100
Q1: Are the two event A={Male} and B={Favor the new gun law} independent?
Q2: How about event A and C={oppose the new gun law}?

43. Disjointness and Independency
The following table shows the political affiliation of voters in a city and their positions on stronger gun control laws:
Party affil
Stronger gun con
Dem
Rep
Total
Favor 60 10
oppose 30
Let A={Dem}, B={oppose}, C={Rep}
Q1: Are A and B disjoint?
Q2: Are A and B independent?

44. Summary to probability rules
Key word
Key idea/ Key conditions
Prob rules
Sum OR
Disjoint /
Mutually exclusive
Pr(A or B) = Pr(A)+Pr(B);
Pr(A and B) = 0.
Not disjoint / mutually exclusive
Pr(A or B) = Pr(A)+Pr(B) -P(A and B)
multiplication
AND
(both)
(ALL)
Independent (with replacement)
Pr(A and B) = Pr(A)٠Pr(B);
Pr(B|A) = Pr(B).
Dependent (without replacement)
Pr(A and B) = Pr(A)٠Pr(B|A)
complement
NOT
at least / at most
Pr( not A) = 1 - Pr(A)

45. Chapter 5 Probability in Our Daily Lives
Section 5.3
Conditional Probability: The Probability of A Given B

46. Conditional Probability
Figure 5.9 Venn Diagram of Conditional Probability of Event A Given Event B. Of the cases in which B occurred, P(A|B) is the proportion in which A also occurred. Question: Sketch a representation of P(B|A). Is P(A|B) necessarily equal to P(B|A)?

47. Conditional Probability
For events A and B, the conditional probability of event A, given that event B has occurred, is:
is read as “the probability of event A, given event B.” The vertical slash represents the word “given”. Of the times that B occurs,
is the proportion of times that A also occurs.
Table for blood type and gender
Find the probability a person’s blood type is A given that they are male.
P(A | Male)=
2. Find the probability a person’s blood type is O given that they are female.
P(O | Female)=
3. Find the probability a person’s blood type is AB given that they are female.
0.034

48. Example 1: Conditional Probability
What was the probability of being audited, given that the income was $1,000,000?
Event A: Taxpayer is audited
Event B: Taxpayer’s income > $1,000,000
Q: Find Pr(A|B)

49. Example 2: Conditional Probability
A study of 5282 women aged 35 or over analyzed the Triple Blood Test to test its accuracy.
A positive test result states that the condition is present.
A negative test result states that the condition is not present.
False Positive: Test states the condition is present, but it is actually absent.
False Negative: Test states the condition is absent, but it is actually present.

50. Example 2: Conditional Probability
Q1: Find the probability of a positive test for a randomly chosen pregnant woman.
Q2: Given that the diagnostic test result is positive, find the estimated probability that Down syndrome truly is present.
P(POS) = 1355/5282 = 0.257

51. Multiplication Rule for Finding P(A and B) Amendment to rule #4
For events A and B, the probability that A and B both occur equals:
also

52. Example: Double Faults in Tennis
Rafael Nadal – 2010 men’s champion in the Wimbledon tennis tournament.
He made 68% of his first serves
He faulted on his first serve the other 32% of the time
Given that he made a fault with his first serve, he made a fault on his second serve only 9% of the time
Q: what is the probability that he makes a double fault when he serves?
Solution: Let F1 be the event that Nadal makes a fault with the first serve, and let F2 be the event that he makes a fault with the second serve.
The event that Nadal makes a double fault is “F1 and F2.”
Nadal makes a double fault in just under 3% of his service points.

53. Independent Events Defined Using Conditional Probabilities
Two events A and B are independent if the probability that one occurs is not affected by whether or not the other event occurs.
Events A and B are independent if:
, or equivalently,
If events A and B are independent,

54. SUMMARY: Checking for Independence
To determine whether events A and B are independent:
Is ?
Is ?
If any one of these is true, the other two are also true, and the events A and B are independent.

55. Example: Down Syndrome Again
The diagnostic blood test for Down syndrome:
POS = positive result NEG = negative result
D = Down Syndrome DC = Unaffected
Status
POS
NEG
Total D
0.009
0.001
0.010 Dc
0.247
0.742
0.990
0.257
0.743
1.000
Are the events POS and D independent or dependent?
Is ?
The events POS and D are dependent.

56. SUMMARY: Probability of an Event Rule #1
The Complement of an Event: P(Not A) Rule # P(Ac) = 1 – P(A).
Probability of the Union of Two Events Rule #3
Addition Rule:
For the union of two events,P(A or B) = P(A) + P(B) – P(A and B)
If the events are disjoint P(A and B) = 0, so P(A or B) = P(A) + P(B)
Multiplication Rule for Finding P(A and B) Amendment to rule #4
For the intersection of two independent events, A and B, P(A and B) = P(A) x P(B)

57. Summary to probability rules
Key word
Key idea/ Key conditions
Prob rules
Sum OR
Disjoint /
Mutually exclusive
Pr(A or B) = Pr(A)+Pr(B);
Pr(A and B) = 0.
Not disjoint / mutually exclusive
Pr(A or B) = Pr(A)+Pr(B) -P(A and B)
multiplication
AND
(both)
(ALL)
Independent
(with replacement)
Pr(A and B) = Pr(A)٠Pr(B);
Pr(B|A) = Pr(B).
Dependent
(without replacement)
Pr(A and B) = Pr(A)٠Pr(B|A)
complement
NOT
at least / at most
Pr( not A) = 1 - Pr(A)
Conditional Probability
Given that
N/A
Pr(B|A): Let Event A had happened, reconstruct a new sample space S*. Then inside S*, find the prob of B.