Download probability basics, part 1

Chapters 13, 14, Part 1
Probability Basics
• Laws of Probability
• Odds and Probability
• Probability Trees
Warm-up
• Your friend is tossing a fair coin and obtains 5
heads in a row.
• After the 5th head she proclaims “I’m
really due for a tail on the next toss.”
• Explain why this thinking is incorrect.
Warm-up (2)
• Non-existent “Law of Averages”
– She is not “due” for a tail just because she has
5 heads in a row. (short-run behavior)
• Law of Large Numbers
– As we repeat a random process over and over,
the proportion of times that an event occurs
settles down to one number. We call this
number the probability. (long-run behavior is a
long time! Infinite)
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
Probability
•Formal study of uncertainty
•The engine that drives Statistics
• Primary objectives of Chapters 13, 14:
1. use the rules of probability to calculate
appropriate measures of uncertainty.
2. Learn the probability basics so that we
can do Statistical Inference
Probability Considerations
•
Your favorite basketball team has the ball and trails by 2 points with
little time remaining in the game. Should your team attempt a gametying 2-pointer or go for a buzzer-beating 3-pointer to win the game?
(This situation has often been used in Microsoft job interviews).
• After a touchdown should a coach kick the extra point or go for two?
• The traffic light at Western Blvd/Gorman St is set to be green for
Western Blvd 65% of the time. The light has been red the previous 7
times you drove through that intersection on Western Blvd. Today you
are “due” for a green light.
• On 4th down should your favorite football team punt or try for the first
down?
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 - 2
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 - 2
• 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
deck of 52 cards.
S = {all 52 cards}
Laws of Probability
1. 0  P ( A)  1, for any event A
2. P (F )  0, P ( S )  1
Laws of Probability - 2
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(Tail) = 0.5

Venn diagram:
Sample space made up of an event
A and its complement 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
Laws of Probability - 3
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 - 4
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
22
Laws of Probability - 5
Multiplication Rule
6. For two independent events A and B
P(A and B) = P(A) × P(B)
Note: assuming events are independent
doesn’t make it true.
Multiplication Rule
• The probability that you encounter a green light at
the corner of Dan Allen and Hillsborough is 0.35, a
yellow light 0.04, and a red light 0.61. What is the
probability that you encounter a red light on both
Monday and Tuesday?
• It’s reasonable to assume that the color of the light
you encounter on Monday is independent of the
color on Tuesday. So
P(red on Monday and red on Tuesday) =
P(red on Monday) × P(red on Tuesday) = 0.61 × 0.61
= 0.3721
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 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)
6. For two independent events A and B
P(A and B) = P(A) × P(B)
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: college students
Suppose 56% of all students live on campus, 62% of all
students purchase a campus meal plan and 42% do both.
Question: what is the probability that a randomly selected
student either lives OR eats on campus.
•L=
{student lives on campus}
• M = {student purchases a meal plan}
P(a student either lives or eats on campus)
= P(L or M) = P(L) + P(M) - P(L and M)
=0.56 + 0.62 – 0.42
= 0.76
Chapter 13 (cont.)
Odds and Probabilities
Probability Trees
ODDS AND
PROBABILITIES
World Series Odds
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
Chapter 13 (cont.)
Probability Trees
A Graphical Method for
Complicated Probability
Problems
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
clinical
reliability
clinical
reliability
Multiply
branch probs
Question 1 Answer
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 Answer
Answer
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
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.