Chapter 14 Probability Basics• Laws of Probability• Odds and
Probability• Probability Trees
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: 23No. of people 23 30 40 60Probability .507.706.891.994
Probability
•Formal study of uncertainty•The engine that drives Statistics
• Primary objective of lecture unit 4:
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
Tomorrow's Weather
A phenomenon is random if individual
outcomes are uncertain, but there is
nonetheless a regular distribution of
outcomes in a large number of repetitions.
Randomness and probabilityRandomness ≠ chaos
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 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).
First series of tossesSecond series
The probability of heads is 0.5 = the proportion of times you get heads in many repeated trials.
4.1 The Laws of Probability
1. Relative frequencyevent probability = x/n, where x=# of occurrences of event of interest, n=total # of observations– Coin, die tossing; nuclear power
plants?
• Limitationsrepeated observations not practical
Approaches to Probability
Approaches to Probability (cont.)
2. Subjective probabilityindividual assigns prob. based on personal experience, anecdotal evidence, etc.
3. Classical approachevery 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 coin2. Draw 1 card from a standard
deck of cards3. 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 FCertain event: S
Examples1. Toss a coin once
S = {H, T}; A = {H}, B = {T}2. Toss a die once; count dots on
upper faceS = {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 adeck of 52 cards.S = {all 52 cards}
Laws of Probability
1)(,0)(.2
event any for ,1)(0 1.
SPP
AAP
Coin Toss Example: S = {Head, Tail}Probability of heads = 0.5Probability 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
Probability rules (cont’d)
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: 23No. of people 23 30 40 60Probability .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}
502.498.1)'(1)(
498.365
343
365
363
365
364)'(
:23365
363
365
364)'(:3
APAPso
AP
people
APpeople
Unions: , orIntersections: , and
AÇB
AÈB
Mutually Exclusive (Disjoint) Events
• Mutually exclusive ordisjoint events-no outcomesfrom S in common
A and B disjoint: A B=
A and B not disjoint
AÈB
AÇB
Venn Diagrams
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)
20
For any two events A and B
P(A or B) = P(A) + P(B) - P(A and B)P(A or B) = P(A) + P(B) - P(A and B)
A
B
P(A) =6/13
P(B) =5/13
P(A and B) =3/13
A or B
+_
P(A or B) = 8/13
General Addition Rule
Laws of Probability (cont.)
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,
thenP(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
Color Brown Red Yellow Green Orange Blue
Probability 0.3 0.2 0.2 0.1 0.1 ?
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:
What is the probability that an M&M chosen at random is blue?
What is the probability that a random M&M is any of red, yellow, or orange?
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
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
Chapter 14 (cont.)
Odds and ProbabilitiesProbability Trees
ODDS AND PROBABILITIES
World Series Odds
From probability to oddsFrom 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), thenP(UNC wins)=3/4
P(UNC does not win)=1/4
Chapter 14 (cont.) Probability Trees
A Graphical Method for Complicated Probability Problems
Example: Prob. of playing pro baseball
Data shows the following probabilities concerning high school baseball players:• 6.1% of hs players go on to play at the college level• 9.4% of hs players that play in college go on to play
professionally• 0.2% of hs players that do not compete in college play
pro baseball What is the probability that a high school
baseball ultimately plays professional baseball?
Probability Tree Approach
A probability tree is a useful way to visualize this problem and to find the desired probability.
Q1: probability that a high school baseball ultimately plays professional baseball?
P(play college and play professionally)
= .061*.094= .005734
Play coll 0.061
Does not play coll 0.939
Play prof. .094
HS BB Player
.906
Play prof. .002
Does not Play prof. .998
P(do not play college and play professionally)
= .939*.002= .001878
P(high school baseball player plays professional baseball) = .005734 + .001878 = .007612
Question 2 Given that a high school player
played professionally, what is the probability he played in college?
.005734.7533
.005734 .001878
P(play college and play professionally)
= .061*.094= .005734
Play coll 0.061
Does not play coll 0.939
Play prof. .094
HS BB Player
.906
Play prof. .002
Does not Play prof. .998
P(do not play college and play professionally)
= .939*.002= .001878
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 TreeMultiply
branch probsclinical reliability
clinical reliability
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
Answertwo 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 = .01593Question 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 .9992. 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.
examples1. P(A)=.3, P(B)=.4; if A and B are
mutually exclusive events, then P(AB)=?
A B = , P(A B) = 02. 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