STAT131 Probability from Tables

STAT131 Probability from Tables W3 L2 Anne Porter Email: alp@uow.edu.au, Room: 15.132/133

An example: STAT131 Mobile phone use

Row margins Column Margins An example: STAT131 Mobile phone use How might we better represent this?

Bar chart: frequency of male and females by category of use What does this graph reveal?

Alternatively the data could be plotted as follows. Which provides the best comparison? Is this really the graph required?

Are these the graphs required? • While both are ways of showing a comparison both basically reveal that there are more males than females. • Neither graph is good for comparing the pattern of usage for males and females What else might we do?

Example 1: proportions What proportion of students is male? What proportion of students is female?

Example 1: proportions 121/168 What proportion of students is male? What proportion of students is female? 47/168

Example 2: row marginal probabilities What is the probability that a randomly selected student will be male? What is the probability that a randomly selected student will be female? The sum of the row marginal probabilities is ?

Example 2: row marginal probabilities What is the probability that a randomly selected student will be male? What is the probability that a randomly selected student will be female? P(Male)=P(M) = 121/168 P(female) = 47/168 1 The sum of the row marginal probabilities is ?

Example 3: column marginal probabilities • What is the probability that a randomly selected student has • No Phone? • An Optus Provider? • A Telstra Provider? • A Vodofone provider?

Example 3: column marginal probabilities • What is the probability that a randomly selected student has • No Phone? • An Optus Provider? • A Telstra Provider? • A Vodofone provider? P(Phone1)=P(No Phone) = 63/168 P(Phone2)=P(Optus) = 55/168 P(Telstra) = 26/168 P(Vodofone) = 24/168 The sum of the column marginal probabilities is ? 1

Example 4: Joint Probabilities • What is the probability of a student selected being • Being male and no mobile? • Being male and optus? • Being male and telstra? • Being male and Vodofone? • Being female and no mobile? • Being female and optus? • Being female and telstra? • Being female and vodofone

Example 4: Joint Probabilities • What is the probability of a student selected being • Being male and no mobile? • Being male and optus? • Being male and telstra? • Being male and Vodofone? • Being female and no mobile? • Being female and optus? • Being female and telstra? • Being female and vodofone P(M and None) = 48/168 P(M and telstra) = 17/168 P(M and Vodo) = 20/168 P(F and none) = 15/168 ...and all the joint probabilities will sum to ? 1

Example 5a: Conditional probabilities • Given that the student is female what is the probability of the student • Having no mobile? • Uses Optus? • Uses telstra? • Uses Vodophone?

Example 5a: Conditional probabilities • Given that the student is female what is the probability of the student • Having no mobile? • Uses Optus? • Uses telstra? • Uses Vodophone? P(No mobile|female) = 15/47 P(Optus|female) = 19/47 P(Telstra|female) = 9/47 P(Vodo|female)= 4/47

Example 5b: Conditional probabilities • Given that the student is male what is the probability of the student • Having no mobile? • Uses Optus? • Uses telstra? • Uses Vodophone?

Example 5b: Conditional probabilities • Given that the student is male what is the probability of the student • Having no mobile? • Uses Optus? • Uses telstra? • Uses Vodophone? P(No mobile|male) = 48/121 P(Optus|male) = 36/121 P(Telstra|male) = 17/121 P(Vodo|male)= 20/121

Based on the answers provided does it appear that the pattern of phone ownership appears to be the same for males and females? 0.32 0.40 0.19 0.09 • P(No mobile|female) = 15/47 • P(Optus|female) = 19/47 • P(Telstra|female) = 9/47 • P(Vodo|female)= 4/47 • P(No mobile|male) = 48/121 • P(Optus|male) = 36/121 • P(Telstra|male) = 17/121 • P(Vodo|male)= 20/121 0.40 0.30 0.14 0.17 What do we need to do?

Example 7: Comparing with a table of percentages • What does this reveal? • How might we better show the comparison?

Example 7: Comparing the pattern of usage for males and females

Example 8: Conditional probabilities What do the following mean? P(Male|No mobile)? P(Male|Optus)? You MUST learn to read the notation correctly

Example 8: Conditional probabilities What do the following mean? P(Male|No mobile)? P(Male|Optus)? What is the probability of being a male if the student has Optus. Answer 48/63 What is the probability the student will have male given they are Optus. Answer 36/55 You MUST learn to read the notation correctly

8. Relative Frequency of Sex given Carrier How do we get the following percentages?

9. Questions about only those who use a carrier What must we do?

Given we have a female and they are a phone user • What is the probability that they • Use Optus? • Use Telstra • Use Vodophone? P(optus|(male and user)= P(Telstra|(male and user)= P(Vodophone|(male and user)= Is the pattern of usage the same for males and females?

Given we have a female and they are a phone user • What is the probability that they • Use Optus? • Use Telstra • Use Vodophone? 19/32 9/32 4/32 P(optus|(male and user)= P(Telstra|(male and user)= P(Vodophone|(male and user)= 36/73 17/73 20/73 Is the pattern of usage the same for males and females?

For users of mobile phones does the carrier appear to be independent of sex? Males appear ro use Vodophone more than females and optus less. But is this just the sample or a pattern evident in the population?

When the pattern of use appears thesame then we have independence We’ve done the sums: how come it is hard to think of what the data are saying?

Can you see THE tree Any one probability rule is simple but put them together...

You need to see THE tree Any one rule is simple but put them together...

How come it is hard to know what the data are saying? • There are a multitude of possible questions • We need appropriate techniques for comparing, visualising ..

Formal definition: Conditional probability P(A|B)=P(A and B)/P(B) • Many of the questions done may be reworked from the definition

Multiplication Rule From P(A|B)=P(A and B)/P(B) By rearrangement we have P(A and B)=P(A|B)xP(B) Equally we can have P(A and B)=P(B|A)xP(A)

Mathematical Independence • If knowing that event B occurred gives no information as to whether A is more or less likely then P(A|B)=P(A) then the events are said to be independent. • If the events are not independent they are dependent. • If P(A and B)=P(A)xP(B) then A and B are independent.

Independence • Independence can be forced through the design of an experiment • For example successive tosses of a coin • When events are independent we can use • P(A and B)= P(A|B) x P(B) • and if indpendent P(A and B) = P(A) x P(B)

When the pattern of use appears identical then we have mathematical independence When we use samples to comment on a population,if the patterns are similar we have independence. How similar we meet later.

Graphical representations

Activity 6 p11: Probability using a Tree Diagram • What is the probability of getting three heads in three independent, fair tosses of a coin. Use a tree diagram to enumerate the possible outcomes and confirm your answer.

P(H)=0.5 P(H)=0.5 P(T)=0.5 P(H)=0.5 P(T)=0.5 P(T)=0.5 Toss 1 Toss 2 P(HHH)=?

P(H)=0.5 P(H)=0.5 P(T)=0.5 P(H)=0.5 P(H)=0.5 HTH P(T)=0.5 P(T)=0.5 HTT P(H)=0.5 THH P(H)=0.5 P(T)=0.5 THT P(T)=0.5 P(T)=0.5 P(H)=0.5 TTH P(T)=0.5 TTT Toss 1 Toss 2 Toss 3 P(HHH)=? HHH HHT P(3H) =1/8

Or using multiplication rule for independent events =1/8 =0.125 X P(H)=0.5 HHH P(H)=0.5 X P(T)=0.5 HHT P(H)=0.5 P(H)=0.5 HTH P(T)=0.5 P(T)=0.5 HTT P(H)=0.5 THH P(H)=0.5 P(T)=0.5 THT P(T)=0.5 P(T)=0.5 P(H)=0.5 TTH P(T)=0.5 TTT Toss 1 Toss 2 Toss 3

P(H)=0.5 P(H)=0.5 P(T)=0.5 P(H)=0.5 P(H)=0.5 HTH P(T)=0.5 P(T)=0.5 HTT P(H)=0.5 THH P(H)=0.5 P(T)=0.5 THT P(T)=0.5 P(T)=0.5 P(H)=0.5 TTH P(T)=0.5 TTT Toss 1 Toss 2 Toss 3 P(2HT)=? HHH HHT P(2HT)=3/8

P(2HT)=? P(H)=0.5 HHH P(H)=0.5 X X P(T)=0.5 HHT P(H)=0.5 X P(H)=0.5 HTH P(T)=0.5 X P(T)=0.5 HTT X P(H)=0.5 THH P(H)=0.5 X P(T)=0.5 THT P(T)=0.5 P(T)=0.5 P(H)=0.5 TTH P(T)=0.5 TTT Toss 1 Toss 2 Toss 3

Law of Total Probability It is through summing the joint occurrences (H&H&T) and (H&T&H) and (T&H&H) that one finds the probability two heads. That is we use the Law of Total Probability. The addition of all these joint probabilities will total 1.

Activity 7 p12:Using Tree diagrams • John has coronary artery disease. He and his doctor must decide between medical management of the disease and coronary bypass surgery. Because John has been quite active, he is concerned about his quality of life as well as length of life. He wants to make the decisian that will maximise the probability of the event A that he survives for 5 years and is able to carry on moderate activity during that time.

Activity 7 cont. • The doctor makes the following probability estimates for patients of John's age and condition: • Under medical management, the probability of A is 0.7 • There is a probability of 0.05 that John will not survive bypass surgery, probability 0.10 that he will survive with serious complications, and probability 0.85 that he will survive the surgery without complications. • Continued next page.

Activity 7 continued • If he survives with complications, the conditional probability of the desired outcome A is 0.73. If there are no serious complications, the conditional probability of A is 0.76. • (a) Draw a tree that summarises this information. • (b) Then calculate the probability of A given that John chooses the surgery. • (c) Does surgery or medical management offer him a better chance of achieving his goal?

STAT131 Probability from Tables