Applied Statistics 270 - Chapter 3

Front 1

The probablity of an event is calculated by

Back 1

summing the probabilities of the sample points in the sample space for A

Front 2

Steps to calculating probabilities for mutually exclusive events
(5)

Back 2

1. define the experiment
2. list the sample points for each variable
3. assign probabilities to the sample points
4. define the set of sample points contained in the event of interest
5. sum the sample point probabilities to get the event probability

Front 3

Example of calculating probability: what is the probability of rolling at least a 9 with a single toss of two dice?
(5)

Back 3

1. The experiment is the single toss of 2 dice
2. The sample points are all the different values that the 2 dice can have. Interested in all that add up to 9, 10, 11, or 12
3. Probabilities of the sample points are: 1/36
4. the set of sample points contained in the event of interest are: (3,6) (4,6) (4,5) (5,6) (5,5) (5,4) (6,6) (6,5) (6,4) (6,3)
5. sum that sample point probabilities to get the event probability:
a. there are 10 points in the event of interest, each having a probability of 1/36. therefore, the event probability is 10/36 = .28

Front 4

Compound event
(2)

Back 4

1. a composition of 2 or more events
2. can be the result of a union or an intersection

Front 5

Union

Back 5

The union of two events A and B is the event that occurs if EITHER A or B or BOTH occur on a single performance of the experiment.

Front 6

The union of events is written

Back 6

A U B

A U B --> all the sample points that belong to A or B or both

Front 7

Intersection

Back 7

The intersection of two events A and B is the event that occurs if BOTH A and B occur on a single performance of the experiment.

Front 8

The intersection of events is written

Back 8

A (upside down U) B

A (upside down U) B consists of all the sample points belonging to BOTH A and B

Front 9

The Additive Rule of Probability

Back 9

The probability of the UNION of events A and B is the sum of the probability of each event minus the probability of their INTERSECTION

Front 10

Formula for the Additive Rule of Probability

Back 10

P(A U B) = P(A) + P(B) - P(A udU B)

Front 11

Mutually exclusive

Back 11

If event A and event B are mutually exclusive, then their union, event C = A U B, looks like two separate circles

Front 12

In other words, Events A and B are mutally exclusive if

Back 12

A udU B contains no sample points

Front 13

For mutually exclusive events,

Back 13

P(A udU B) = 0

Front 14

If B and C are mutually exclusive events,

Back 14

their intersection is empty

Front 15

Example of mutually exclusive events

Back 15

Birthplace - there is nobody who was born in both Elko and Reno

Front 16

Complement of an event

Back 16

1. The complement of an event A, A^c, is the event that A does not occur.
2. A^c consists of all sample points that are not in A

Front 17

Example of complementary events

Back 17

If A is at least 1 head in a toss of 2 coins, A^c is having no heads appear.

A: HH, HT, TH
A^c: TT

Front 18

The Rule of complements

Back 18

The sum of the probabilities of complementary events equals 1

P(A) + P(A^c) = 1
or
P(A) = 1 - P(A^c)

Front 19

Conditional Probability

Back 19

1. Before, we assumed that all "conditions" were defined by the experiment.
2. But sometimes we have additional knowledge about conditions that can affect the outcome of an experiment.
3. You can think of this additional information as a reduction in the sample space that may affect the probability of some events

Front 20

Conditional Probability formula

Back 20

P(A.line.B) = P(A udU B) / P(B)

Front 21

The multiplicative rule and independent events

Back 21

Events A and B are independent if the occurrence of one does not alter the probability of the other

Front 22

If A and B are independent events, the formulas are:
(3)

Back 22

1. P(A.line.B) = P(A)
2. P(B.line.A) = P(B)
3. P(A udU B) = P(A) x P(B)

Front 23

Checking whether B and C are independent events
(2)

Back 23

Check whether
1. P(B.line.C) = P(B)
2. P(B udU C) = P(B) x P(C.line.B)

Front 24

Random sample
(df)

Back 24

n units are selected from a population in such a way that EVERY set of n units in that population has an equal probability of being selected

Front 25

To find out how many random samples there are, we turn to

Back 25

combinatorial mathematics

Front 26

Combinations Rule

Back 26

A sample of n elements is to be drawn from a set of N elements. The number of different samples possible is written:

(N over n) = N! / n!(N-n)!

Front 27

Example of combinations rule

Back 27

Consider the task of choosing 2 marines from a platoon of 4 to send on a dangerous mission. How many different selections can be made?

Combinations Rule:
(4 over 2) = 4! / (2!)(4-2)!
= 4! / (2!)(2!)
= 6

Front 28

Finding a random sample

Back 28

The U.S. Bureau of the Census regularly samples the population for demographic information such as income, family size, employment, and marital status. Suppose the Bureau plans to sample 1,000 households in a town that has a total population of 53,432 households.

How could the Bureau use the random # table in Appendix A or a computer to generate the sample? Select the first 10 households in the sample.

Front 29

How could the Bureau use the random # table in Appendix A or a computer to generate the sample?
(5)

Back 29

my own words: bureau gives a # to every unit. then uses a computer to generate a sample of random #s


1. Number the households from 00,001 to 53,432
2. using the random number table, select a column at random, say column 8.
3. the first 10 numbers in column 8 will be the first 10 households to sample.
4. Select the next 990 between 00,001 and 53,432 (just discard the ones outside your range)
5. the sample will consist of the 1,000 households corresponding to the 1,000 different numbers

Front 30

To do random sampling of people in the household, use a

Back 30

kish grid

(could take next birthday to get a random person to sample)