Transcript Chapter 14 From Randomness to Probability

Chapters 14 & 15 Probability
math2200
Random phenomenon
• In a random phenomenon we know what
could happen, but we don’t know which
particular outcome did or will happen.
• Example: the color of the traffic light at a
particular intersection.
Accumulated percentage
Day
[1,]
[2,]
[3,]
[4,]
[5,]
[6,]
[7,]
[8,]
[9,]
[10,]
Light
G
G
G
R
G
G
G
R
R
G
fraction
1.0000000
1.0000000
1.0000000
0.7500000
0.8000000
0.8333333
0.8571429
0.7500000
0.6666667
0.7000000
0
20
40
60
# of outcomes
80
100
0.5
0.6
0.7
fraction
0.8
0.9
1.0
Probability
• Each occasion upon which we observe a random
phenomenon is called a trial.
• We observe the value of the random phenomenon and
call it an outcome.
• The collection of all possible outcomes forms the sample
space.
• When we combine outcomes, the resulting combination
is an event. An event occurs if the outcome of the trial is
in this event.
• Trials are independent if the outcome of one trial does
not influence the outcome of the others.
• The long-run relative frequency of an event is called the
probability of the event.
Example: Two coins
• Trial: flip two coins at the same time
• Outcome: one of the four combinations of heads
(H) or tails (T)
• Sample space: all possible outcomes
– S = {HH,TT,HT,TH}
• Events:
–
–
–
–
Event A : two coins give the same results
{ HH,TT }
Event B : two coins give different results
{ HT,TH }
The Law of Large Numbers (LLN)
• The Law of Large Numbers (LLN)
says that the long-run relative
frequency of repeated
independent events gets closer
and closer to a single value. We
call the value the probability of
the event.
• this definition of probability is
often called empirical probability
because it is based on repeatedly Jacob Bernoulli
observing the event’s outcome, .
CAUTION: The Nonexistent Law of
Averages
• The LLN says nothing about short-run
behavior.
• Relative frequencies even out only in the
long run (infinitely long, in fact).
• The so called Law of Averages (that an
outcome of a random event that hasn’t
occurred in many trials is “due” to occur)
doesn’t exist at all.
Modeling probability: Example
• When probability was first studied, a group
of French mathematicians looked at
games of chance in which all the possible
outcomes were equally likely.
– It’s equally likely to get any one of six
outcomes from the roll of a fair die.
– It’s equally likely to get heads or tails from the
toss of a fair coin.
Probability (cont.)
• If Event A is made up of several equally
likely outcomes.
# of outcomes in A
P(A) =
# of possible outcomes
• Example:
the probability of drawing a non-face card
(A-10) from a deck?
Personal Probability
• In everyday speech, when we express a
degree of uncertainty without basing it on
long-run relative frequencies, we are
stating subjective or personal probabilities.
• Personal probabilities don’t display the
kind of consistency that we will need
probabilities to have.
Formal Probability
1. A probability must be a number between
0 and 1.
For any event A, 0 ≤ P(A) ≤ 1.
Formal Probability (cont.)
2. Probability Assignment Rule:
– The probability of the set of all possible
outcomes of a trial must be 1.
P(S) = 1 (S denotes the set of all possible
outcomes.)
Formal Probability (cont.)
rules of computations
3. Complement Rule:

The set of outcomes that are not in the event A is
called the complement of A, denoted AC.The
probability of an event occurring is 1 minus the
probability that it doesn’t occur: P(A) = 1 – P(AC)
Formal Probability - Notation
Notation alert:
• In this class we use the notation P(A or
B) and P(A and B).
• In other situations, you might see the
following:
– P(A  B) instead of P(A or B)
– P(A  B) instead of P(A and B)
Formal Probability (cont.)
4. Addition Rule:
– Events that have no outcomes in common
(and, thus, cannot occur together) are called
disjoint (or mutually exclusive).
Formal Probability (cont.)
4. Addition Rule (cont.):
– For two disjoint events A and B, the
probability that one or the other occurs is the
sum of the probabilities of the two events.
P(A or B) = P(A) + P(B), provided that A
and B are disjoint.
For more events:
P(A or B or C) = P(A) + P(B) + P(C), where A, B
and C are mutually exclusive
Be careful about natural language!
– Often, “or” in our natural language has an
exclusive meaning as in “Would you like the
steak or the vegetarian entrée?”.
– In this class, when we ask for the probability
that A or B occurs, we mean A or B or both.
– P(A or B but not both) = P(A or B) – P(A and B)
= P(A) + P(B) – 2 * P(A and B)
The General Addition Rule (cont.)
• General Addition Rule:
– For any two events A and B,
P(A or B) = P(A) + P(B) – P(A and B)
– P(A or B but not both) = P(A or B) – P(A and B)
= P(A) + P(B) – 2 * P(A and B)
General addition rule
For three events:
P(A or B or C) = P(A) + P(B) + P(C)- P(A and B) –
P(A and C) – P(B and C) + P(A and B and C)
Formal Probability
5. Multiplication Rule (cont.):
– For two independent events A and B, the
probability that both A and B occur is the
product of the probabilities of the two events.
P(A and B) = P(A) x P(B), provided that A
and B are independent.
P(A and B and C) = P(A) x P(B) x P(C) , if A
and B and C are mutually independent
Formal Probability (cont.)
5. Multiplication Rule (cont.):
– Two independent events A and B are not
disjoint, provided the two events have
probabilities greater than zero:
Multiplication rule
• For two independent events A and B, the
probability that both A and B occur is the product
of the probabilities of the two events.
P(A and B) = P(A) * P(B), if A and B are
independent
• For more events,
P(A and B and C) = P(A) * P(B) * P(C) , if A and B
and C are mutually independent
Example: M&M
• In 2001, Masterfoods decided to add
another color to the standard color lineup
of brown, yellow, red, orange, blue and
green. To decide which color, they
surveyed kids in nearly every country and
asked them to vote for purple, pink and
teal.
Example: M&M
• In Japan, the result is somehow different.
– 38% for pink
– 36% for teal
– 16% purple
• What is the probability that a Japanese
M&M’s survey respondent selected at
random preferred either pink or teal?
Example (Cont’)
• If we pick two respondents at random,
what’s the probability that they both said
pink or teal?
Example (cont’)
• If we pick three respondents at random, what’s
the probability that at least one preferred purple?
Solution: Let A be the event that at least one
preferred purple. Therefore AC is the event that
non of the three prefer purple.
P(A)=1-P(AC) = 1- (0.84)^3=40.73%
Solution 2: P(A) =P (Exactly one respondent
likes purple) + P (Exactly two like purple)+ P
(Exactly three like purple)= 3*0.16*0.84*0.84
+3*0.16*0.16*0.84 + 0.16^3 = 40.73%
What can go wrong?
• Be aware of probabilities that don’t add up
to 1.
• Don’t add probabilities of events if they are
NOT disjoint.
• Don’t multiply probabilities of events if they
are NOT independent.
• Independent ≠ Disjoint
Independent ≠ Disjoint
• Disjoint events cannot be independent! Well,
why not?
– Since we know that disjoint events have no outcomes
in common, knowing that one occurred means the
other didn’t.
– Thus, the probability of the second occurring changed
based on our knowledge that the first occurred.
– It follows that the two events are not independent.
• A common error is to treat disjoint events as if
they were independent, and apply the
Multiplication Rule for independent events—
don’t make that mistake.
Example
• Police report that 78% of drivers stopped
on suspicion of drunk driving are given a
breath test, 36% a blood test, and 22%
both tests. What is the probability that a
randomly selected suspect is given
– a blood test or a breath test?
– neither test?
• A = {the suspect is given a breath test}
• B = {the suspect is given a blood test}
• We know that
– P(A) = 0.78
– P(B) = 0.36
– P(A and B) = 0.22
• P(A or B) = P(A) + P(B) – P(A and B)
=0.78 + 0.36 – 0.22 = 0.92
• P(A or B but NOT both)
– P(A or B) – P(A and B) = 0.92 – 0.22 = 0.70
– P(A or Bc) + P(B or Ac)
• P(Ac and Bc) = 1 – P(A or B)
Checking disjoint
• Are giving a suspect a blood test and a
breath test mutually exclusive?
– This is to see whether P(A and B) = 0
– In this case, P(A and B) = 0.22
– So, not mutually exclusive
Checking independence
• Are giving a suspect a blood test and a
breath test independent?
– P(A and B) = P(A) * P(B)?
Example
• Two psychologists surveyed 478 children.
• They asked the students whether their
primary goal was
– to get good grades
– to be popular
– or to be good at sports
• Purpose of the study
– Did boys and girls at this age have similar
goals?
Conditional distributions
grades
Boy
Girl
popular
sports
Total
117 (51.54%) 50 (22.03%) 60 (26.43%) 227
grades
popular
sports
Total
130 (51.79%)
91 (36.25%)
30 (11.95%)
251
Conditional probability
• the probability of an event from a
conditional distribution is noted P(B|A) and
pronounce it “the probability of B given A.”
• A probability that takes into account a
given condition is called a conditional
probability.
• Conditional probability of B given A
P(B| A)  P(A and B)
P(A)
• Note: P(A) cannot equal 0, since we know
that A has occurred.
The General Multiplication Rule
• Rearranging the equation in the definition
for conditional probability, we get the
General Multiplication Rule:
– For any two events A and B,
P(A and B) = P(A) x P(B|A)
or
P(A and B) = P(B) x P(A|B)
Independence
• Definition: Events A and B are independent
whenever P(B|A) = P(B) (or P(A|B) = P(A)) .
• Using the general multiplication rule, we can see
that it’s equivalent to P(A and B) = P(A) * P(B) .
P(A and B) = P(A) * P(B|A) = P(A) * P(B) ( or
P(A) and B) = P(B)*P(A|B) = P(B) *P(A) )
Boy
Girl
Total
grades
117
130
247
popular
50
91
141
sports
60
30
90
Total
227
251
478
• Trial: select a student at random (equally
likely) and asks about his/her goal.
• Sample space: 478 students
• What is the probability the selected
student is a girl? (251/478 = 0.525)
• P (girl and popular) = ?
• P (sports) = 90/478 = 0.188
Conditional probability
• Given that the selected student is a girl,
what is the probability the selected
student’s goal is sports? P (sports | girl) =
P (sports and girl) / P (girl) = (30/478) /
(251/478) =30/251= 0.12
• P (sports | boy) = 60/227 = 0.264
• P (sports) ≠ P (sports | girl) + P (sports |
boy)
• P (girl | sports) = ?
Example
• Police report that 78% of drivers stopped on
suspicion of drunk driving are given a breath test,
36% a blood test, and 22% both tests.
• Are giving a DWI suspect a blood test and a
breath test independent?
P(B|A) = P(A and B) / P(A) = 0.22 / 0.78 = 0.28
P(B) = 0.36
Therefore P(B|A) ≠ P(B), the event of blood test
and the event of breath test are not
independent
Drawing Without Replacement
• Sampling without replacement means that once
one object is drawn it doesn’t go back into the
pool. (It does not matter to large population).
• Suppose that 12 rooms left when it is time for
you and your friend to draw. What is the
probability that both of you get rooms in Gold
Hall? (1/22)
– Three are in Gold Hall
– Four in Silver Hall
– Five in Wood Hall
Example: Binge drinking
• A study by Harvard School of Public Health, for
college students
– 44% engage in binge drinking
– 37% drink moderately
– 19% abstain entirely
• Meanwhile, another study
– Among binge drinkers, 17% involved in alcoholrelated automobile accident
– Among students who drink moderately , only 9% have
been involved
Tree Diagrams
• A tree diagram helps us
think through conditional
probabilities by showing
sequences of events as
paths that look like
branches of a tree.
Reversing the conditioning
• If we know a student has had an alcoholrelated accident, what is the probability
that the student is a binge drinker?
– P (binge | accident) = ?
– P (binge and accident) / P (accident)
– P (accident) = P (binge and accident) +
P (moderate and accident) + P (abstain and
accident) = 0.075+ 0.033 + 0 = 0.108
Bayes’ rule
• What is P(B|A), how do we get P(A|B)
P(B|A) = P(A and B)/P(A) = P(A|B)*P(B)/P(A)
P(A) = P(A and B) + P(A and Bc)
= P(A|B)*P(B) + P (A | Bc)* P(Bc)
P(B|A) = P(A|B)*P(B)/[P(A|B)* P (B) + P(A| Bc)* P(Bc).
We need P(B), P(A|B) and P(A | Bc)
What can go wrong?
• Be aware of probabilities that don’t add up
to 1.
• Don’t add probabilities of events if they are
NOT disjoint.
• Don’t multiply probabilities of events if they
are NOT independent.
• Independent ≠ Disjoint