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Probability

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Probability

Formal study of uncertainty

The engine that drives statistics

- Nothing in life is certain
- We gauge the chances of successful outcomes in business, medicine, weather, and other everyday situations such as the lottery (recall the birthday problem)

- For most of human history, probability, the formal study of the laws of chance, has been used for only one thing: gambling

- Nobody knows exactly when gambling began; goes back at least as far as ancient Egypt where 4-sided “astragali” (made from animal heelbones) were used

Rule 1: Let Caesar win IV

out of V times

- The Roman emperor Claudius (10BC-54AD) wrote the first known treatise on gambling.
- The book “How to Win at Gambling” was lost.

- Relative frequency
event probability = x/n,

where x=# of occurrences of event of interest, n=total # of observations

- Coin, die tossing; nuclear power plants?
- Limitations
repeated observations not practical

- Subjective probability
individual assigns prob. based on personal experience, anecdotal evidence, etc.

- Classical approach
every possible outcome has equal probability (more later)

- Experiment: act or process that leads to a single outcome that cannot be predicted with certainty
- Examples:
1.Toss a coin

2.Draw 1 card from a standard deck of cards

3.Arrival time of flight from Atlanta to RDU

- Sample space: all possible outcomes of an experiment. Denoted by S
- Event: any subset of the sample space S;
typically denoted A, B, C, etc.

Simple event: event with only 1 outcome

Null event: the empty set F

Certain event: S

1.Toss a coin once

S = {H, T}; A = {H}, B = {T} simple events

2.Toss a die once; count dots on upper face

S = {1, 2, 3, 4, 5, 6}

A=even # of dots on upper face={2, 4, 6}

B=3 or fewer dots on upper face={1, 2, 3}

3.P(A’ ) = 1 - P(A)

For an event A, A’ is the complement of A; A’ is everything in S that is not in A.

S

A'

A

- What is the smallest number of people you need in a group so that the probability of 2 or morepeople having the same birthday is greater than 1/2?
- Answer: 23
No. of people23304060

Probability.507.706.891.994

- A={at least 2 people in the group have a common birthday}
- A’ = {no one has common birthday}

S

AÇB

A

B

AÈB

- Mutually exclusive events-no outcomes from S in common

A Ç B = Æ

S

A

B

Addition Rule for Disjoint Events:

4. If A and B are disjoint events, then

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

- 5. For two independent events A and B
P(A B) = P(A) × P(B)

General Addition Rule

6. For any two events A and B

P(A B) = P(A) + P(B) – P(A B)

S

AÇB

A

B

- S = {1, 2, 3, 4, 5, 6}
- A = even # appears = {2, 4, 6}
- B = 3 or fewer = {1, 2, 3}
- P(A È B) = P(A) + P(B) - P(A Ç B)
=P({2, 4, 6}) + P({1, 2, 3}) - P({2})

= 3/6 + 3/6 - 1/6 = 5/6

- 1. 0 P(A) 1 for any event A
- 2. P() = 0, P(S) = 1
- 3. P(A’) = 1 – P(A)
- 4. If A and B are disjoint events, then
P(A B) = P(A) + P(B)

- 5. If A and B are independent events, then
P(A B) = P(A) × P(B)

- 6. For any two events A and B,
P(A B) = P(A) + P(B) – P(A B)

Probability Models

The Equally Likely Approach

(also called the Classical Approach)

- If an experiment has N outcomes, then each outcome has probability 1/N of occurring
- If an event A1 has n1 outcomes, then
P(A1) = n1/N

We Need Efficient Methods for Counting Outcomes

- A student wishes to commute to a junior college for 2 years and then commute to a state college for 2 years. Within commuting distance there are 4 junior colleges and 3 state colleges. How many junior college-state college pairs are available to her?

- junior colleges: 1, 2, 3, 4
- state colleges a, b, c
- possible pairs:
(1, a) (1, b) (1, c)

(2, a) (2, b) (2, c)

(3, a) (3, b) (3, c)

(4, a) (4, b) (4, c)

- junior colleges: 1, 2, 3, 4
- state colleges a, b, c
- possible pairs:
(1, a) (1, b) (1, c)

(2, a) (2, b) (2, c)

(3, a) (3, b) (3, c)

(4, a) (4, b) (4, c)

4 junior colleges

3 state colleges

total number of possible

pairs = 4 x 3 = 12

- junior colleges: 1, 2, 3, 4
- state colleges a, b, c
- possible pairs:
(1, a) (1, b) (1, c)

(2, a) (2, b) (2, c)

(3, a) (3, b) (3, c)

(4, a) (4, b) (4, c)

In general, if there are n1 ways

to choose the first element of

the pair, and n2 ways to choose

the second element, then the

number of possible pairs is

n1n2. Here n1 = 4, n2 = 3.

- NCAA Basketball Tournament: how many ways can the “bracket” be filled out?
- How many games?
- 2 choices for each game
- Number of ways to fill out the bracket:
263 = 9.2 × 1018

- Earth pop. about 6 billion; everyone fills out 1 million different brackets
- Chances of getting all games correct is about 1 in 1,000

- Pollsters minimize lead-in effect by rearranging the order of the questions on a survey
- If Gallup has a 5-question survey, how many different versions of the survey are required if all possible arrangements of the questions are included?

- There are 5 possible choices for the first question, 4 remaining questions for the second question, 3 choices for the third question, 2 choices for the fourth question, and 1 choice for the fifth question.
- The number of possible arrangements is therefore
5 4 3 2 1 = 120

- Factorial Notation:
n!=12 … n

- Examples
1!=1; 2!=12=2; 3!= 123=6; 4!=24;

5!=120;

- Special definition: 0!=1

- Calculator:
non-graphing: x ! (second function)

graphing: bottom p. 9 T I Calculator Commands

(math button)

- Excel:
Paste: math, fact

- 20! = 2.43 x 1018
- 1,000,000 seconds?
- About 11.5 days
- 1,000,000,000 seconds?
- About 31 years
- 31 years = 109 seconds
- 1018 = 109 x 109
- 31 x 109 years = 109 x 109 = 1018 seconds
- 20! is roughly the age of the universe in seconds

A B C D E

- How many ways can we choose 2 letters from the above 5, without replacement, when the order in which we choose the letters is important?
- 5 4 = 20

- Calculator
non-graphing: nPr

- Graphing
p. 9 of T I Calculator Commands

(math button)

- Excel
Paste: Statistical, Permut

A B C D E

- How many ways can we choose 2 letters from the above 5, without replacement, when the order in which we choose the letters is not important?
- 5 4 = 20 when order important
- Divide by 2: (5 4)/2 = 10 ways

From the numbers 1 through 20,

choose 6 different numbers.

Write them on a piece of paper.

Prior to Jan. 1, 2009

After Jan. 1, 2009

- How large is 195,249,054?
- $1 bill and $100 bill both 6” in length
- 10,560 bills = 1 mile
- Let’s start with 195,249,053 $1 bills and one $100 bill …
- … and take a long walk, putting down bills end-to-end as we go

… still plenty of bills remaining, so continue from …

… still plenty of bills remaining, so continue from…

… still plenty of bills remaining, so continue from …

… still plenty of bills remaining, so continue from …

… still plenty of bills remaining, so …

Still have ~ 5,000 bills left!!

- Remember: one of the bills you put down is a $100 bill; all others are $1 bills
- Your chance of winning the lottery is the same as bending over and picking up the $100 bill while walking the route blindfolded.

Probability Trees

A Graphical Method for Complicated Probability Problems

- V={person has HIV}; CDC: P(V)=.006
- +: test outcome is positive (test indicates HIV present)
- -: test outcome is negative
- clinical reliabilities for a new HIV test:
- If a person has the virus, the test result will be positive with probability .999
- If a person does not have the virus, the test result will be negative with probability .990

- What is the probability that a randomly selected person will test positive?

- A probability tree is a useful way to visualize this problem and to find the desired probability.

clinical reliability

clinical reliability

Multiply

branch probs

clinical reliability

clinical reliability

- What is the probability that a randomly selected person will test positive?
- P(+) = .00599 + .00994 = .01593

- If your test comes back positive, what is the probability that you have HIV?
(Remember: we know that if a person has the virus, the test result will be positive with probability .999; if a person does not have the virus, the test result will be negative with probability .990).

- Looks very reliable

Answer

two sequences of branches lead to positive test; only 1 sequence represented people who have HIV.

P(person has HIV given that test is positive) =.00599/(.00599+.00994) = .376

- Question 1:
- P(+) = .00599 + .00994 = .01593
- Question 2: two sequences of branches lead to positive test; only 1 sequence represented people who have HIV.
P(person has HIV given that test is positive) =.00599/(.00599+.00994) = .376

- We have a test with very high clinical reliabilities:
- If a person has the virus, the test result will be positive with probability .999
- If a person does not have the virus, the test result will be negative with probability .990

- But we have extremely poor performance when the test is positive:
P(person has HIV given that test is positive) =.376

- In other words, 62.4% of the positives are false positives! Why?
- When the characteristic the test is looking for is rare, most positives will be false.

1. P(A)=.3, P(B)=.4; if A and B are mutually exclusive events, then P(AB)=?

A B = , P(A B) = 0

2. 15 entries in pie baking contest at state fair. Judge must determine 1st, 2nd, 3rd place winners. How many ways can judge make the awards?

15P3 = 2730