1. Probability Part2 (5.3-5.4-5.5-5.6)
Chapter 5
Probability
Part2 ( )

2. 5.3 multiplication rule Definitions
Two events E and F are independent if the occurrence of event E in a probability experiment does not affect the probability of event F.
Two events are dependent if the occurrence of event E in a probability experiment affects the probability of event F.

3. multiplication rule:
Example: Woman will have two children. Assume the probability the birth results in boy is Then probability of a boy followed by a girl is (0.51)(0.49) = About a 25% chance a woman will have a boy and then a girl.

4. More examples:
Suppose you toss three coins. What is the probability of getting three tails?
1/2 x 1/2 x 1/2 = 1/8
Find the probability that a 100-year flood will strike a city in two consecutive years
1 in 100 x 1 in 100 = 0.01 x 0.01 =

5. EXAMPLE Computing Probabilities of Independent Events
A manufacturer of exercise equipment knows that 10% of their products are defective. They also know that only 30% of their customers will actually use the equipment in the first year after it is purchased. If there is a one-year warranty on the equipment, what proportion of the customers will actually make a valid warranty claim?
We assume that the defectiveness of the equipment is independent of the use of the equipment. So,

7. = P (1st survives and 2nd survives and 3rd survives and 4th survives)
EXAMPLE Illustrating the Multiplication Principle for Independent Events
The probability that a randomly selected female aged 60 years old will survive the year is % according to the National Vital Statistics Report, Vol. 47, No What is the probability that four randomly selected 60 year old females will survive the year?
P(all four survive)
= P (1st survives and 2nd survives and 3rd survives and 4th survives)
= P(1st survives) . P(2nd survives) . P(3rd survives) . P(4th survives)
= ( ) ( ) ( ) ( ) =

8. “at least” probability
Probability of At Least Once
What is the probability of something happening at least once?
P(at least one event E in n trials) =
1 - [P(not E in one trial)]n

9. EXAMPLE 1 Computing “at least” Probabilities
What is the probability that a region will experience at least one 100-year flood during the next 55 years?
Probability of a flood is 1/100. Probability of no flood is 99/100.
P(at least one flood in 55 years) = = 0.425

10. EXAMPLE 2 Computing “at least” Probabilities
The probability that a randomly selected female aged 60 years old will survive the year is % according to the National Vital Statistics Report, Vol. 47, No What is the probability that at least one of 500 randomly selected 60 year old females will die during the course of the year?
P(at least one dies) = 1 – P(none die)
= 1 – P(all survive)
= 1 – =

11. Summary of probability rules:

12. Conditional Probability
Section Conditional Probability and the
General Multiplication Rule
Conditional Probability
The notation P(F | E) is read “the probability of event F given event E”.
It is the probability of an event F given the occurrence of the event E.

13. EXAMPLE
Suppose that a single six-sided die is rolled. What is the probability that the die comes up 4? Now suppose that the die is rolled a second time, but we are told the outcome will be an even number. What is the probability that the die comes up 4?
First roll: S = {1, 2, 3, 4, 5, 6} P(4) = 1/6
Second roll: S = {2, 4, 6} P(4 | it’s even)=1/3
(probability of a 4 given result will be even)

15. EXAMPLE Murder Victims
In 2005, 19.1% of all murder victims were between the ages of 20 and 24 years old. Also in 2005, 16.6% of all murder victims were 20 – 24 year old males. What is the probability that a randomly selected murder victim in 2005 was male given that the victim is years old?
= 0.869

16. Example: ELISA HIV TEST:
ELISA is a test for AIDS used to screen donated blood for hiv. (See )
If a blood sample tests positive, then one is concerned that it is contaminated with the AIDS virus.
The test is not always correct, though. If the blood does have the AIDS virus, then there is 99% chance it will say so.
If the blood does not have aids, there is still a 6% chance it will have a positive result. About 1% of the blood samples are thought to be contaminated.

17. Here's a simplified scenario for 1,010,000 people donating blood and the results on their blood:

18. Randomly pick one person from these 1,001,000
Randomly pick one person from these 1,001, Fill in the blanks below:
The chance the person has AIDS is____________________
The chance the person tests positive given the person has AIDS is _________
The chance the person tests positive given the person does not have AIDS is_____
The chance the person has AIDS given the person does not test positive is__________
The chance the person has AIDS given the person tests positive is____________
Are "having AIDS" and "testing Positive" independent? _______

19. Section 5.5 Combinatorial counting rule
What is the number of all possible combinations when r items are taken from a total of n?
Also:
Answer:
The symbol nCr represents the number of combinations of n distinct objects taken r at a time, where r < n.

20. Reminder:

21. Example:
How many possible ways are there to pick 3 m & m’s with different colors from a bag that has 9 different colors? (pink,blue,red,yellow,green,gray,brown,purple, orange)
Answer: 9C3=
There are 84 combinations of 3 different colors.

22. The United States Senate consists of 100 members
The United States Senate consists of 100 members. In how many ways can 4 members be randomly selected to attend a luncheon at the White House?
Answer: 100C4

23. Section 5.6 Rule of Total Probability
EXAMPLE
At a university
55% of the students are female and 45% are male
15% of the female students are business majors
20% of the male students are business majors
What percent of students, overall, are business majors?

24. Answer
The percent of the business majors in the university contributed by females:
55% of the students are female
15% of those students are business majors
Thus 15% of 55%, or 0.55 • 0.15 = or 8.25% of the total student body are female business majors
Contributed by males:
In the same way, 20% of 45%, or 0.45 • 0.20 = .09 or 9% are male business majors

25. Better Answer: Use of Tree Diagram
Female
Male
0.55
0.45
0.55•0.15
0.55•0.85
0.45•0.20
0.45•0.80
Business
Not Business
0.0825
0.0900
Total =
0.4675
0.3600

26. This is an example of the Rule of Total Probability
P(Bus) = 55% • 15% + 45% • 20%
= P(Female) • P(Bus | Female)
+ P(Male) • P(Bus | Male)
This rule is useful when the sample space can be divided into two (or more) disjoint parts

27. EXAMPLE (Rule of Total Probability)
In a particular town
30% of the voters are Republican
30% of the voters are Democrats
40% of the voters are independents
This is a partition of the voters into three sets
There are no voters that are in two sets (disjoint)
All voters are in one of the sets (covers all of S)
For a particular issue
90% of the Republicans favor it
60% of the Democrats favor it
70% of the independents favor it
These are the conditional probabilities
E = {favor the issue}
The above probabilities are P(E | political party)

28. The total proportion of votes who favor the issue
0.3 • • • 0.7 = 0.73
So 73% of the voters favor this issue

29. What percent of business majors
Reverse problem:
(Also can use Bayes’s Rule – see textbook)
We could turn this problem around
We were told the percent of female students who are business majors
We could also ask:
What percent of business majors
are female?

30. Answer:
If we chose a random business major, what is the probability that this student is female?
A1 = Female student
A2 = Male student
E = business major
We want to find P(A1 | E): the probability that the student is female (A1) given that this is a business major (E)

31. (continued)
We know that 8.25% of the students are female business majors.
We know that 9% of the students are male business majors
Choosing a business major at random is choosing one of the 17.25%
The probability that this student is female is 8.25% / 17.25% = 47.83%

32. Using: Bayes’s Rule it’s the same calculation
Bayes’s Rule for a partition into two sets (n = 2)
P(A1) = .55, P(A2) = .45
P(E | A1) = .15, P(E | A2) = .20
We know all of the numbers we need