Probability
•Formal study of uncertainty
•The engine that drives statistics
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 (recall the birthday problem)
History
• For most of human history, probability, the
formal study of the laws of chance, has
been used for only one thing: gambling
History (cont.)
• 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
History (cont.)
• The Roman emperor Claudius (10BC-54AD)
wrote the first known treatise on gambling.
• The book “How to Win at Gambling” was
lost.
Rule 1: Let Caesar win IV
out of V times
Approaches to Probability
• 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.)
• Subjective probability
individual assigns prob. based on personal
experience, anecdotal evidence, etc.
• 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.
Simple event: event with only 1 outcome
Null event: the empty set F
Certain event: S
Examples
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}
Laws of Probability
1. 0  P( A)  1, for any event A
2. P(F )  0, P( S )  1
Laws of Probability (cont.)
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
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 and Intersections
S
A
B
A
A
Mutually Exclusive Events
• Mutually exclusive events-no outcomes
from S in common
A = 
S
A
B
Laws of Probability (cont.)
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)
Laws of Probability (cont.)
General Addition Rule
6. For any two events A and B
P(A  B) = P(A) + P(B) – P(A  B)
P(AB)=P(A) + P(B) - P(A B)
S
A
A
B
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 B) = P(A) + P(B) - P(A B)
=P({2, 4, 6}) + P({1, 2, 3}) - P({2})
= 3/6 + 3/6 - 1/6 = 5/6
Laws of Probability: Summary
•
•
•
•
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)
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
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,4cjunior colleges
• possible pairs: 3 state colleges
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, 2, 3,
In 4
general, if there are n1 ways
to choose the first element of
• state colleges a, b, cthe 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:
Paste: math, fact
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
31 x 109 years = 109 x 109 = 1018 seconds
20! is roughly the age of the universe in
seconds
Permutations
ABCDE
• 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
Paste: Statistical, Permut
Combinations
ABCDE
• 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 101 Powerball Lottery
From the numbers 1 through 20,
choose 6 different numbers.
Write them on a piece of paper.
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!
North Carolina Powerball Lottery
Prior to Jan. 1, 2009
After Jan. 1, 2009
5 from 1 - 55:
5 from 1 - 59:
55!
 3, 478, 761
5!50!
59!
 5, 006, 386
5!54!
1 from 1 - 42 (p'ball #):
1 from 1 - 39 (p'ball #):
42!
 42
1!41!
39!
 39
1!38!
3, 478, 761*42 
146,107, 962
5, 006, 386*39 
195, 249, 054
Visualize 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 endto-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
• Your chance of winning the lottery is the
same as bending over and picking up the
$100 bill while walking the route
blindfolded.
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)
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
Example: AIDS Testing
•
•
•
•
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:
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
Probability Tree
clinical
reliability
clinical
reliability
Multiply
branch probs
Question 1 Answer
• What is the probability that a randomly
selected person will test positive?
• 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 Answer
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
Summary
• 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
•
Recap
We have a test with very high clinical reliabilities:
1.
2.
•
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.
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