- 57 Views
- Uploaded on
- Presentation posted in: General

Chapter Four

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Chapter Four

Elementary Probability Theory

Focus Points

- Assign probabilities to events.
- Explain how to law of large numbers relates to relative frequencies.
- Apply basic rules of probability in everyday life.
- Explain the relationship between statistics and probability.

- Probability is a numerical measurement of likelihood of an event.
- The probability of any event is a number between zero and one.
- Events with probability close to one are more likely to occur.
- Events with probability close to zero are less likely to occur.

If A represents an event,

P(A)

represents the probability of A.

Event A is certain to occur

Event A is certain not to occur

- Intuition
- Relative frequency
- Equally likely outcomes

- Incorporates past experience, judgment, or opinion.
- Is based upon level of confidence in the result
- Example: “I am 95% sure that I will attend the party.”

Probability of an event =

the fraction of the time that the event occurred in the past =

f

n

where f = frequency of an event

n = sample size

Example of Probability

as Relative Frequency

If you note that 57 of the last 100 applicants for a job have been female, the probability that the next applicant is female would be:

- No one result is expected to occur more frequently than any other.

When tossing a coin, the probability of getting head=

In the long run, as the sample size increases and increases, the relative frequencies of outcomes get closer and closer to the theoretical (or actual) probability value.

- Any random activity that results in a definite outcome

- A collection of one or more outcomes of a statistical experiment or observation

- An outcome of a statistical experiment that consists of one and only one of the outcomes of the experiment

- The set of all possible distinct outcomes of an experiment
- The sum of all probabilities of all simple events in a sample space must equal one.

1, 2, 3, 4, 5, 6

- P(even number) =
- P(result less than five) =
- P(not getting a two) =

- the event that A does not occur
- Notation for the complement of event A:

- P(event A does not occur) =
P(Ac) = 1 – P(A)

- So, P(A) + P(Ac) = 1

If the probability that it will snow today is 30%,

P(It will not snow) = 1 – P(snow) =

1 – 0.30 = 0.70

- Probability makes statements about what will occur when samples are drawn from a known population.
- Statistics describes how samples are to be obtained and how inferences are to be made about unknown populations.

- The occurrence (or non-occurrence) of one event does not change the probability that the other event will occur.

- P(A and B) = P(A) •P(B)

- If events are dependent, the fact that one occurs affects the probability of the other.
- P(A, givenB) equals the probability that event A occurs, assuming that B has already occurred.

- P(A and B) = P(A) •P(B, given A)
- P(Aand B) = P(B) •P(A, given B)

- For independent events:
P(A and B) = P(A) •P(B)

- For any events:
P(A and B) = P(A) •P(B, given A)

P(A and B) = P(B) •P(A, given B)

When choosing two cards from two separate decks of cards, find the probability of getting two fives.

P(two fives) =

P(5 from first deck and 5 from second) =

When choosing two cards from a deck without replacement, find the probability of getting two fives.

P(two fives) =

P(5 on first draw and 5 on second) =

- And means both events occur together.
- Or means that at least one of the events occur.

For any eventsA and B,

P(A or B) =

P(A) + P(B) – P(A and B)

P(5 ) + P(red) – P(5 and red) =

When choosing a card from an ordinary deck, the probability of getting a five or a six:

P(5 ) + P(6) – P(5 and 6) =

- Events that are disjoint, cannot happen together.

For any mutually exclusive events A and B,

P(A or B) = P(A) + P(B)

P(4 or 6) =

P(male and college grad) = ?

P(male and college grad) =

P(male or college grad) = ?

P(male or college grad) =

P(male, given college grad) = ?

P(male, given college grad) =

- Tree Diagram
- Multiplication Rule of Counting
- Permutations
- Combinations

- a visual display of the total number of outcomes of an experiment consisting of a series of events

Experiment of rolling two dice, one after the other and observing any of the six possible outcomes each time .

Number of paths = 6 x 6 = 36

For any series of events, if there are

n1possible outcomes for event E1

and n2 possible outcomes for event E2, etc.

then there are

n1X n2X···X nm possible outcomes

for the series of events E1 through Em.

In the past, an area code consisted of 3 digits,

the first of which could be any digit from 2 through 9.

The second digit could be only a 0 or 1.

The last could be any digit.

How many different such area codes were possible?

8·2 ·10 = 160

In how many different ways could four items be arranged in order from first to last?

4 ·3 ·2·1 = 24

- n! is read "n factorial"
- n! is applied only when n is a whole number.
- n! is a product of n with each positive counting number less than n

120

- 5! = 5 • 4 • 3 • 2 • 1 =
- 3! = 3 • 2 • 1 =

6

- 1! = 1
- 0! = 1

24

4! =

10! =

6! =

15! =

3,628,800

720

1.3077 x 1012

- A Permutation is an arrangement in a particular order of a group of items.
- There are to be no repetitions of items within a permutation.

How many different permutations of the letters a, b, c are possible?

Solution: There are six different permutations:

abc, acb, bac, bca, cab, cba.

How many different two-letter permutations of the letters a, b, c, d are possible?

Solution: There are twelve different permutations:

ab, ac, ad, ba, ca, da, bc, bd, cb, db, cd, dc.

The number of ways to arrange in order n distinct objects, taking them r at a time, is:

where n and r are whole numbers and n>r.

P3, 3 = _______ P4, 2 = _______

P6, 2 = __________P8, 3 = _______

P15, 2 = _______

6

12

30

336

210

A teacher has chosen eight possible questions for an upcoming quiz. In how many different ways can five of these questions be chosen and arranged in order from #1 to #5?

Solution: P8,5 =

= 8• 7 • 6 • 5 • 4 = 6720

A combination is a grouping

in no particular order

of items.

The number of combinations of n objects taken r at a time is:

where n and r are whole numbers and n>r.

10

35

C5, 3 = C7, 3 =

C3, 3 = C15, 2 =

C6, 2 =

1

105

15

A teacher has chosen eight possible questions for an upcoming quiz. In how many different ways can five of these questions be chosen if order makes no difference?

Solution: C8,5 =

= 56

- If the experiment consists of a series of stage with various outcomes, use the multiplication rule or a tree diagram.

If the outcomes consist of ordered subgroups of r outcomes taken from a group of n outcomes use the permutation rule.

If the outcomes consist of non-ordered subgroups of r items taken from a group of n items use the combination rule.