1. Chapter 15: Probability Rules!

2. A “Refresher” Each trial generates an outcome
An event is any set or collection of outcomes
The collection of all possible outcomes is called the sample space, denoted S

3. Events When outcomes are equally likely, each has a probability of 1/k
𝑃 𝐴 = 𝑐𝑜𝑢𝑛𝑡 𝑜𝑓 𝑎𝑙𝑙 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑖𝑛 𝐴 𝑐𝑜𝑢𝑛𝑡 𝑜𝑓 𝑎𝑙𝑙 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠
This ONLY works when outcomes are equally likely!!

4. Not Disjoint?
What’s the probability of rolling a number less than 3 or an even number on a die? A = # less than 3 = {1, 2} B = even number = {2, 4, 6} A ∩ B = 2, so the events are not disjoint General Addition Rule - does not require disjoint events 𝑃 𝐴∪𝐵 =𝑃 𝐴 +𝑃 𝐵 −𝑃 𝐴∩𝐵
Subtract out the duplication

5. Just Checking
Chapter 1 suggested that you sample some pages of this book at random to see whether they held a graph or other data display. They actually did just that and drew a representative sample and found the following: 48% of pgs had some kind of data display 27% of pgs had an equation, and 7% of pgs had both a data display & an equation

6. Just Checking Display these results in a Venn diagram
What is the probability that a randomly selected sample page has neither a data display nor an equation?
What is the probability that a randomly selected sample page has a data display but no equation?

7. Step-by-Step Example
Police report that 78% of drivers stopped on suspicion of drunk driving are given a breath test, 36% a blood test, and 22% both tests. What is the probability that a randomly selected DWI suspect is given
a test?
a blood test or a breath test, but not both?
neither test?

8. Step-by-Step Example Plan – define the events we’re interested in
Plot – make a Venn diagram
Let A = {suspect given a breath test}
Let B = {suspect given a blood test}
𝑃 𝐴 =0.78
𝑃 𝐵 =0.36
𝑃 𝐴∩𝐵 =0.22
So,
𝑃 𝐴∩ 𝐵 𝐶 =0.78−0.22=0.56
𝑃 𝐵∩ 𝐴 𝐶 =0.36−0.22=0.14
𝑃 𝐴 𝐶 ∩ 𝐵 𝐶 =1− =0.08
0.56
0.22
0.14
A B
0.08

9. Step-by-Step Example 𝑃 𝐴∪𝐵 =𝑃 𝐴 +𝑃 𝐵 −𝑃 𝐴∩𝐵
Question 1:
Show
𝑃 𝐴∪𝐵 =𝑃 𝐴 +𝑃 𝐵 −𝑃 𝐴∩𝐵
= – 0.22 = 0.92
Tell
92% of all suspects are given a test.

10. Step-by-Step Example 𝑃 𝐴 𝑜𝑟 𝐵 𝑏𝑢𝑡 𝑛𝑜𝑡 𝑏𝑜𝑡ℎ =𝑃 𝐴∪𝐵 −𝑃 𝐴∩𝐵
Question 2:
Show
𝑃 𝐴 𝑜𝑟 𝐵 𝑏𝑢𝑡 𝑛𝑜𝑡 𝑏𝑜𝑡ℎ =𝑃 𝐴∪𝐵 −𝑃 𝐴∩𝐵
= 0.92 – 0.22 = 0.70
Tell
70% of the suspects get exactly one of the tests.

11. Step-by-Step Example 𝑃 𝑛𝑒𝑖𝑡ℎ𝑒𝑟 𝑡𝑒𝑠𝑡 =1−𝑃 𝑒𝑖𝑡ℎ𝑒𝑟 𝑡𝑒𝑠𝑡 = 1 – 0.92 = 0.08
Question 3:
Show
𝑃 𝑛𝑒𝑖𝑡ℎ𝑒𝑟 𝑡𝑒𝑠𝑡 =1−𝑃 𝑒𝑖𝑡ℎ𝑒𝑟 𝑡𝑒𝑠𝑡
= 1 – 0.92 = 0.08
Tell
Only 8% of the suspects get no test.

12. It Depends…
Goals
Grades
Popular
Sports
Total
Boy
117 50 60
227
Girl
130 91 30
251
247
141 90
478
Sex
Two psychologists surveyed 478 children in grades 4, 5, and 6 in elementary school in Michigan. They stratified their sample, drawing roughly 1/3 from rural, 1/3 from suburban, and 1/3 from urban schools. Among other questions, they asked the students whether their primary goal was to get good grades, to be popular, or to be good at sports. One question of interest was whether boys and girls at this age had similar goals. Here’s the contingency table of the survey’s data.

13. Some Examples
What’s the probability that a student selected at random is a girl?
What’s the probability that a student selected at random is a girl whose goal is to be popular?
What’s the probability that a student selected at random has a goal to be popular given we have selected a girl?
𝑃 𝑔𝑖𝑟𝑙 = =0.525
𝑃 𝑔𝑖𝑟𝑙∩𝑝𝑜𝑝𝑢𝑙𝑎𝑟 = =0.190
𝑃 𝑝𝑜𝑝𝑢𝑙𝑎𝑟|𝑔𝑖𝑟𝑙 = =0.363

14. Conditional Probability
When we want the probability of an event from a conditional distribution, we write P(B|A) and pronounce it “the probability of B given A)”
A probability that takes into account a given condition such as this is called a conditional probability
𝑃 𝐵|𝐴 = 𝑃 𝐴∩𝐵 𝑃 𝐴

15. The General Multiplication Rule
When A and B are independent events,
𝑃 𝐴∩𝐵 =𝑃 𝐴 ×𝑃 𝐵
The General Multiplication Rule (works when you don’t have independent events):
Using the two previous formulas, we can conclude events A and B are independent when
𝑃 𝐴∩𝐵 =𝑃 𝐴 ×𝑃 𝐵|𝐴
𝑃 𝐵|𝐴 =𝑃 𝐵

16. Independent ≠ Disjoint
Disjoint events cannot be independent. Given two disjoint events: 1. you get an A on the midterm 2. you get a B on the midterm They’re disjoint because they have no outcomes in common. Suppose you DID get an A on the midterm. Now what is the probability that you got a B on the exam? You can’t get both grades, so independence doesn’t even make sense here.

17. Tables and Conditional Probability
Another option to using a Venn diagram is to create a contingency table for data: Example (from earlier): 78% of drivers stopped on suspicion of drunk driving are given a breath test, 36% a blood test, and 22% both tests
Breath Test
Yes No
Total
0.22
0.36
0.78
1.00
0.14
Blood Test
0.56
0.08
0.64
0.22

18. Disjoint? Independent?
Let’s take another look at the drunk driving situation. Police report that 78% of drivers are given a breath test, 36% a blood test, and 22% both.
Are giving a DWI suspect a blood test and a breath test mutually exclusive?
Are giving the two tests independent?

19. Question 1:
Are giving a DWI suspect a blood test and a breath test mutually exclusive? 𝑃 𝐴∩𝐵 =0.22. Since 22% of the suspects are given both tests, 𝑃 𝐴∩𝐵 ≠0. Therefore, a breath test and a blood test are not disjoint events.
Breath Test
Yes No
Total
0.22
.14
0.36
.56
.08
.64
0.78
.22
1.00
Blood Test

20. Question 2:
Are giving the two tests independent? 𝑃 𝐵 =0.36 𝑃 𝐵|𝐴 = 𝑃 𝐴∩𝐵 𝑃 𝐴 = ≈0.28 𝑃 𝐵|𝐴 ≠𝑃 𝐵 Overall, 36% of the drivers get blood tests, but only 28% of those who get a breath test do. Since suspects who get a breath test are less likely to have a blood test, the two events are not independent.
Breath Test
Yes No
Total
0.22
.14
0.36
.56
.08
.64
0.78
.22
1.00
Blood Test

21. Just Checking
Remember our sample of pages in this book from this chapter’s first Just Checking…? We found that 48% of pages had a data display, 27% of pages had an equation, and 7% of pages had both a data display and an equation.
Make a contingency table for the variables display and equation
What is the probability that a randomly selected sample page with an equation also had a data display?
Are having an equation and having a data display disjoint events?
Are having an equation and having a data display independent events?

22. “Without Replacement”
There are twelve rooms available in 3 different dormitories for college housing. Gold Hall (3 rooms) is very desirable with spacious, deluxe rooms. Silver Hall (4 rooms) has a great location on campus, but the rooms aren’t as nice. Wood Hall (5 rooms) has a bad location with really small room. Rooms are drawn at random. What’s the probability that the next two people will get Gold Hall? 𝑃 𝐺𝑜𝑙𝑑∩𝐺𝑜𝑙𝑑 =𝑃 𝑔𝑜𝑙𝑑 ×𝑃 𝑔𝑜𝑙𝑑 = 3 12 × 2 11 =0.045
Once the first person chooses a dorm, the pool of rooms changes for the future trials

23. Tree Diagrams
According to a study by the Harvard School of Public Health, 44% of college students engage in binge drinking, 37% drink moderately, and 19% abstain entirely. Another study, published in the American Journal of Health Behavior, finds that among binge drinkers aged 21 to 34, 17% have been involved in an alcohol-related automobile accident, while among nonbingers of the same age, only 9% have been involved in such accidents. What’s the probability that a randomly selected college student will be a binge drinker who has had an alcohol-related car accident?

24. Tree Diagrams 𝑃 𝑏𝑖𝑛𝑔𝑒∩𝑎𝑐𝑐𝑖𝑑𝑒𝑛𝑡 =0.0748 Binge and Accident Accident
.44×.17=.0748
.17
Binge
.83
Binge and None
None
.44×.83=.3652
.44
Accident
Moderate and Accident
Moderate
.37
.09
.37×.09=.0333
.91
None
Moderate
and None
.37×.91=.3367
.19
Abstain
Accident
Abstain and Accident
.19×0=0 1
Abstain and None
None
.19×1=.19
𝑃 𝑏𝑖𝑛𝑔𝑒∩𝑎𝑐𝑐𝑖𝑑𝑒𝑛𝑡 =0.0748

25. Reversing the Conditioning
What’s the probability the student is a binge drinker given he has an alcohol-related accident? Does the tree diagram help? 𝑃 𝑏𝑖𝑛𝑔𝑒|𝑎𝑐𝑐𝑖𝑑𝑒𝑛𝑡 = 𝑃 𝑏𝑖𝑛𝑔𝑒∩𝑎𝑐𝑐𝑖𝑑𝑒𝑛𝑡 𝑃 𝑎𝑐𝑐𝑖𝑑𝑒𝑛𝑡 = =0.694

26. Another Example
Suppose the incidence of tuberculosis is 5 in 10,000 (rare, but none the less a serious disease). The test to detect TB is 99% accurate. What’s the probability you have TB if your test is positive?
Being 99% accurate means a false-positive rate of 1%.
A false negative is much worse than a false positive – let’s assume a false-negative rate of 0.1%

27. TB Testing We need to use a conditional probability here:+
.999
𝑃 + =𝑃 𝑇𝐵∩+ +𝑃 𝑇 𝐵 𝐶 ∩+ = = TB
.001
0.0005 –
0.9995 +
.01
no TB
.99 –
𝑃 𝑇𝐵|+ = 𝑃 𝑇𝐵∩+ 𝑃 +
= =0.047
The chance of having TB even after you test positive is 4.7%.

28. Bayes’ Rule
Important in Statistics and is the foundation of an approach to statistical analysis known as Bayesian Statistics
“Just” a formula for reversing the probability from the conditional probability that you’re originally given
This can be done (often times, more easily) without the formula
𝑃 𝐵|𝐴 = 𝑃 𝐴|𝐵 𝑃 𝐵 𝑃 𝐴|𝐵 𝑃 𝐵 +𝑃 𝐴| 𝐵 𝐶 𝑃 𝐵 𝐶

29. What Can Go Wrong?
Don’t use a simple probability rule where a general rule is appropriate
Don’t find probabilities for samples drawn without replacement as if they had been drawn with replacement
Don’t reverse conditioning naively
Don’t confuse disjoint with independent