Transcript Chapters 13 and 14 powerpoints only

Chapters 13 and 14
•Probability and counting
Birthday Problem
What is the smallest number of people
you need in a group so that the
probability of 2 or more people having the
same birthday is greater than 1/2?
Answer: 23
No. of people
23 30 40 60
Probability
.507 .706 .891 .994
We will solve this problem a few slides later using the
laws of probability
Probability
•Formal study of uncertainty
•The engine that drives
Statistics
• Primary objectives:
1. use the rules of probability to calculate
appropriate measures of uncertainty.
2. Learn the probability basics so that we
can do Statistical Inference
Introduction
Nothing in life is certain
We gauge the chances of successful
outcomes in business, medicine, weather,
and other everyday situations such as the
lottery or the birthday problem
Randomness and
probability
Randomness ≠ chaos
A phenomenon is random if individual
outcomes are uncertain, but there is
nonetheless a regular distribution of
outcomes in a large number of repetitions.
Coin toss
The result of any single coin toss is
random. But the result over many tosses
is predictable, as long as the trials are
independent (i.e., the outcome of a new
coin flip is not influenced by the result of
the previous flip).
The probability of
heads is 0.5 =
the proportion of
times you get
heads in many
repeated trials.
First series of tosses
Second series
Approaches to Probability
1. 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
Approaches to Probability
(cont.)
2. Subjective probability
individual assigns prob. based on personal
experience, anecdotal evidence, etc.
3. Classical approach
every possible outcome has equal
probability (more later)
Basic Definitions
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
Basic Definitions (cont.)
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.
Null event: the empty set F
Certain event: S
Examples
1.
Toss a coin once
S = {H, T}; A = {H}, B = {T}
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. Select 1 card from a
Laws of Probability
1. 0  P( A)  1, for any event A
2. P(F )  0, P( S )  1
Probability rules (cont’d)
Coin Toss Example:
S = {Head, Tail}
Probability of heads = 0.5
Probability of tails = 0.5
3) The complement of any event A is the event
that A does not occur, written as A.
The complement rule states that the probability
of an event not occurring is 1 minus the
probability that is does occur.
P(not A) = P(A) = 1 − P(A)
Tail  = not Tail = Head
P(Tail ) = 1 − P(Head) = 0.5

Venn diagram:
Sample space made up of an event
A and its complementary A , i.e.,
everything that is not A.
Birthday Problem
What is the smallest number of people
you need in a group so that the
probability of 2 or more people having the
same birthday is greater than 1/2?
Answer: 23
No. of people
23 30 40 60
Probability
.507 .706 .891 .994
Example: Birthday Problem
A={at least 2 people in the group have a
common birthday}
A’ = {no one has common birthday}
3 people
23 people
:P ( A') 
364
363

365
365
:
364
363
343
P ( A') 


 . 498
365
365
365
so P ( A )  1  P ( A ' )  1  . 498  . 502
Unions: , or
Intersections: , and
A
A
Mutually Exclusive
(Disjoint) Events
Venn Diagrams
A and B disjoint: A  B=
Mutually exclusive or
disjoint events-no outcomes
from S in common
A
A
A and B not disjoint
Addition Rule for Disjoint
Events
4. If A and B are disjoint events, then
P(A or B) = P(A) + P(B)
Laws of Probability (cont.)
General Addition Rule
5. For any two events A and B
P(A or B) = P(A) + P(B) – P(A and B)
General Addition Rule
For any two events A and B
P(A or B) = P(A) + P(B) - P(A and B)
P(A) =6/13
+
A
P(B) =5/13
_
B
P(A and B) =3/13
P(A or B) = 8/13
A or B
20
Laws of Probability:
Summary
0  P(A)  1 for any event A
P() = 0, P(S) = 1
P(A’) = 1 – P(A)
If A and B are disjoint events, then
P(A or B) = P(A) + P(B)
5. For any two events A and B,
P(A or B) = P(A) + P(B) – P(A and B)
1.
2.
3.
4.
M&M candies
If you draw an M&M candy at random from a bag, the candy will have one
of six colors. The probability of drawing each color depends on the proportions
manufactured, as described here:
Color
Probability
Brown
Red
Yellow
Green
Orange
Blue
0.3
0.2
0.2
0.1
0.1
?
What is the probability that an M&M chosen at random is blue?
S = {brown, red, yellow, green, orange, blue}
P(S) = P(brown) + P(red) + P(yellow) + P(green) + P(orange) + P(blue) = 1
P(blue) = 1 – [P(brown) + P(red) + P(yellow) + P(green) + P(orange)]
= 1 – [0.3 + 0.2 + 0.2 + 0.1 + 0.1] = 0.1
What is the probability that a random M&M is any of red, yellow, or orange?
P(red or yellow or orange)
= P(red) + P(yellow) + P(orange)
= 0.2 + 0.2 + 0.1 = 0.5
Example: toss a fair die
once
S = {1, 2, 3, 4, 5, 6}
A = even # appears = {2, 4, 6}
B = 3 or fewer = {1, 2, 3}
P(A or B) = P(A) + P(B) - P(A and B)
=P({2, 4, 6}) + P({1, 2, 3}) - P({2})
= 3/6 + 3/6 - 1/6 = 5/6
THE RELATIONSHIP
BETWEEN ODDS AND
PROBABILITIES
•World Series Odds
•The odds at the above link are the odds
against a team winning the World
Series, though the author claims they’re
“odds for winning the World Series”
•Odds are frequently a source of
confusion. Odds for? Odds against?
•From probability to odds
•From odds to probability
From Probability to Odds
If event A has
probability P(A), then
the odds in favor of A
are P(A) to 1-P(A). It
follows that the odds
against A are 1-P(A)
to P(A)
If the probability of
an earthquake in
California is .25, then
the odds in favor of
an earthquake are .25
to .75 or 1 to 3. The
odds against an
earthquake are .75 to
.25 or 3 to 1
From Odds to Probability
If the odds in favor of
an event E are a to b,
then
P(E)=a/(a+b)
in addition,
P(E’)=b/(a+b)
If the odds in favor of
UNC winning the
NCAA’s are 3 (a) to 1
(b), then
P(UNC wins)=3/4
P(UNC does not win)=
1/4
Probability Models
The Equally Likely Approach
(also called the Classical
Approach)
Assigning Probabilities
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
Dice
You toss two dice. What is the probability of the outcomes summing to 5?
This is S:
{(1,1), (1,2), (1,3),
……etc.}
There are 36 possible outcomes in S, all equally likely (given fair dice).
Thus, the probability of any one of them is 1/36.
P(the roll of two dice sums to 5) =
P(1,4) + P(2,3) + P(3,2) + P(4,1) = 4 / 36 = 0.111
We Need Efficient Methods
for Counting Outcomes
Product Rule for Ordered
Pairs
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?
Product Rule for Ordered
Pairs
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)
Product Rule for Ordered
Pairs
junior colleges: 1, 2, 3, 4
state colleges a, b, c
4 junior colleges
3 state colleges
possible pairs:
total number of possible
(1, a) (1, b) (1, c)
pairs = 4 x 3 = 12
(2, a) (2, b) (2, c)
(3, a) (3, b) (3, c)
(4, a) (4, b) (4, c)
Product Rule for Ordered
Pairs
junior colleges: 1, In
2,general,
3, 4 if there are n1 ways
to choose the first element of
state colleges a, b,thec pair, and n ways to choose
2
the second element, then the
possible pairs:
number of possible pairs is
(1, a) (1, b) (1, c) n1n2. Here n1 = 4, n2 = 3.
(2, a) (2, b) (2, c)
(3, a) (3, b) (3, c)
(4, a) (4, b) (4, c)
Counting in “Either-Or” Situations
• NCAA Basketball Tournament: how
many ways can the “bracket” be filled
out?
1. How many games?
2. 2 choices for each game
3. 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
Counting Example
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?
Solution
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
Efficient Methods for
Counting Outcomes
Factorial Notation:
n!=12 … n
Examples
1!=1; 2!=12=2; 3!= 123=6; 4!=24;
5!=120;
Special definition: 0!=1
Factorials with calculators
and Excel
Calculator:
non-graphing: x ! (second function)
graphing: bottom p. 9 T I Calculator
Commands
(math button)
Excel:
Insert function: Math and Trig category,
FACT function
Factorial Examples
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
20! is roughly the age of the universe in
seconds
Permutations
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
Permutations (cont.)
5!
5!
5  4  20 
  5 4
(5  2)! 3!
5!
Notation : 5 P2 
 20
(5  2)!
Permutations with
calculator and Excel
Calculator
non-graphing: nPr
Graphing
p. 9 of T I Calculator Commands
(math button)
Excel
Insert function: Statistical, Permut
Combinations
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
Combinations (cont.)
 
5!
5! 5  4 20
 5 C2 



10
(5  2)!2! 3!2! 1 2 2
 
n!
 n Cr 
(n  r )! r!
5
2
n
r
ST 311 Powerball Lottery
From the numbers 1 through 20,
choose 6 different numbers.
Write them on a piece of paper.
And the numbers are ...
16
11
2
10
8
4
wow
scream
Chances of Winning?
Choose 6 numbers from 20, without
replacemen t, order not important.
Number of possibilit ies?
 
20
6
20!
 20 C6 
 38,760
(20  6)!6!
Example: Illinois State
Lottery
Choose 6 numbers from 54 numbers without
replacemen t; order not important
54!
 25,827,165
54 C6 
48!6!
(about 1 second in 10 months)
(1200 ft 2 house, 16.5 million ping pong balls)
North Carolina Powerball
Lottery
Prior to Jan. 1, 2009
5 from 1 - 55:
55!
 3, 478, 761
5!50!
1 from 1 - 42 (p'ball #):
42!
 42
1!41!
3, 478, 761*42 
146,107, 962
After Jan. 1, 2009
5 from 1 - 59:
59!
 5, 006, 386
5!54!
1 from 1 - 39 (p'ball #):
39!
 39
1!38!
5, 006, 386*39 
195, 249, 054
The Forrest Gump Visualization of
Your Lottery Chances
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
Raleigh to Ft. Lauderdale…
… still plenty of bills
remaining, so continue
from …
… Ft. Lauderdale to San
Diego
… still plenty of bills remaining, so continue from…
… San Diego to Seattle
… still plenty of bills remaining, so continue from …
… Seattle to New York
… still plenty of bills remaining, so continue from …
… New York back to
Raleigh
… still plenty of bills remaining, so …
Go around again! Lay a
second path of bills
Still have ~ 5,000 bills left!!
Chances of Winning NC
Powerball Lottery?
Remember: one of the bills you put down
is a $100 bill; all others are $1 bills.
Put on a blindfold and begin walking
along the trail of bills.
Your chance of winning the lottery is the
same as your chance of selecting the
$100 bill if you stop at a random location
along the trail and pick up a bill .
Virginia State Lottery
50!
Pick 5 : 50 C5 
 2,118,760
45!5!
2,118,760  25 C1 
25!
2,118,760 
 52,969000
24!1!
Probability Trees
A Graphical Method for
Complicated Probability
Problems
Probability Tree Example: probability
of playing professional baseball
 6.1% of high school baseball players play college
baseball. Of these, 9.4% will play professionally.
 Unlike football and basketball, high school players can
also go directly to professional baseball without playing
in college…
 studies have shown that given that a high school player
does not compete in college, the probability he plays
professionally is .002.
Question 1: What is the probability that a high school
baseball player ultimately plays professional baseball?
Question 2: Given that a high school baseball player
played professionally, what is the probability he played in
college?
Question 1: What is the probability that a high school
baseball player ultimately plays professional baseball?
Play prof. .094
.061*.094=.005734
Play coll 0.061
.906
HS BB Player
Play prof. .002
Does not play coll
0.939
Does not Play
prof. .998
.939*.002=.001878
P(hs bb player plays professionally)
= .061*.094 + .939*.002
= .005734 + .001878
= .007612
Question 2: Given that a high school baseball player played
professionally, what is the probability he played in college?
Play prof. .094
Play coll 0.061
.906
.061*.094=.005734
P(hs bb player plays professionally)
= .005734 + .001878
= .007612
HS BB Player
Play prof. .002
.939*.002=.001878
Does not play coll
0.939
Does not Play
prof. .998
P(played in college given that played professionally)
.005734
=
 .7533
.007612
Example: AIDS Testing
 V={person has HIV}; CDC: Pr(V)=.006
 P : test outcome is positive (test
indicates HIV present)
 N : test outcome is negative
 clinical reliabilities for a new HIV test:
1. If a person has the virus, the test result will
be positive with probability .999
2. If a person does not have the virus, the test
result will be negative with probability .990
Question 1
What is the probability that a randomly
selected person will test positive?
Probability Tree Approach
A probability tree is a useful way to
visualize this problem and to find the
desired probability.
Probability Tree
clinical
reliability
clinical
reliability
Multiply
branch probs
Question 1: What is the probability that a
randomly selected person will test positive?
Pr( P)  .00599  .00994  .01593
Question 2
 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
Question 2: If your test comes back positive, what is
the probability that you have HIV?
Pr( P)  .00599  .00994  .01593
P(have HIV given that test is positive)
.00599
=
 .376
.00599  .00994
Summary
Question 1:
Pr(P ) = .00599 + .00994 = .01593
Question 2: two sequences of branches
lead to positive test; only 1 sequence
represented people who have HIV.
Pr(person has HIV given that test is positive)
=.00599/(.00599+.00994) = .376
Recap
 We have a test with very high clinical
reliabilities:
1. If a person has the virus, the test result will be
positive with probability .999
2. 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:
Pr(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.
examples
1. P(A)=.3, P(B)=.4; if A and B are mutually
exclusive events, then P(AB)=?
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