CHAPTER 5ProbabilityPart2 (5.3-5.4-5.5-5.6)

1

5.3 MULTIPLICATION RULE

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.

2

DEFINITIONS

3

MULTIPLICATION RULE:

Example: Woman will have two children. Assume the probability the birth results in boy is 0.51. Then probability of a boy followed by a girl is (0.51)(0.49) = 0.2499. 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 years1 in 100 x 1 in 100 = 0.01 x 0.01 = 0.0001

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?

EXAMPLE Computing Probabilities of Independent EventsEXAMPLE Computing Probabilities of Independent Events

We assume that the defectiveness of the equipment is independent of the use of the equipment. So,

defective and used defective used

(0.10)(0.30)

0.03

P P P

The probability that a randomly selected female aged 60 years old will survive the year is 99.186% according to the National Vital Statistics Report, Vol. 47, No. 28. 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)

= (0.99186) (0.99186) (0.99186) (0.99186)

= 0.9678

EXAMPLE Illustrating the Multiplication Principle for Independent EventsEXAMPLE Illustrating the Multiplication Principle for Independent Events

“AT LEAST” PROBABILITY

8

PROBABILITY OF AT LEAST ONCEWhat 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

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) = 1 - 0.9955 = 0.425

EXAMPLE 1 Computing “at least” ProbabilitiesEXAMPLE 1 Computing “at least” Probabilities

The probability that a randomly selected female aged 60 years old will survive the year is 99.186% according to the National Vital Statistics Report, Vol. 47, No. 28. 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 – 0.99186500

= 0.9832

EXAMPLE 2 Computing “at least” ProbabilitiesEXAMPLE 2 Computing “at least” Probabilities

SUMMARY OF PROBABILITY RULES:

Section 5.4 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.

EXAMPLEEXAMPLE

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)

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 20 - 24 years old?

EXAMPLE Murder VictimsEXAMPLE Murder Victims

male and 20 24male 20 24

20 24

PP

P

0.166

0.191

= 0.869

EXAMPLE: ELISA HIV TEST: ELISA is a test for AIDS used to screen donated blood

for hiv. (See http://www.stat.ucla.edu/cases/elisa/ )

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.

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

Randomly pick one person from these 1,001,000. Fill in the blanks below:

1. The chance the person has AIDS is____________________ 2. The chance the person tests positive given the person has

AIDS is _________3. The chance the person tests positive given the person does

not have AIDS is_____

4. The chance the person has AIDS given the person does not test positive is__________

5. The chance the person has AIDS given the person tests positive is____________

6. Are "having AIDS" and "testing Positive" independent? _______

Section 5.5 Combinatorial counting rule

Answer:

The symbol nCr represents the number of combinations of n distinct objects taken r at a time, where r < n.

What is the number of all possible combinations when r items are taken from a total of n?

Also:

Reminder:

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.

EXAMPLE:

84123

789

123123456

123456789

!3!6

!9

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

Section 5.6 Rule of Total Probability

EXAMPLEEXAMPLE

At a university55% of the students are female and 45% are male15% of the female students are business majors20% of the male students are business majors

What percent of students, overall, are business majors?

AnswerAnswer

● 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 = 0.0825 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

Better Answer: Use of Tree DiagramBetter Answer: Use of Tree Diagram

Business

Not Business

Business

Not Business

Female

Male

0.55

0.45

0.55•0.15

0.55•0.85

0.45•0.20

0.45•0.80

0.0825

0.0900

Total = 0.1725

0.4675

0.3600

• 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

● 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)

EXAMPLE (Rule of Total Probability)EXAMPLE (Rule of Total Probability)

● 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)

● The total proportion of votes who favor the issue

0.3 • 0.9 + 0.3 • 0.6 + 0.4 • 0.7 = 0.73

● So 73% of the voters favor this issue

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 majorsare female?

5-2

9

Reverse problem: (Also can use Bayes’s Rule – see textbook)

● 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)

Answer:Answer:

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 is8.25% / 17.25% = 47.83%

(continued)(continued)

)A|E(P)A(P)A|E(P)A(P)A|E(P)A(P

)E|A(P i2211

11

● 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

Using: Bayes’s RuleUsing: Bayes’s Rule5

-32

4783

090008250825

204515551555

2211

11

.

...

......

)A|E(P)A(P)A|E(P)A(P)A|E(P)A(P