Transcript Chapter 6

Try to estimate the likelihood of a result by
actually observing the random phenomenon
many times and calculating the relative
frequency of the results.
2. Develop a probability model and use it to
calculate a theoretical answer.
3. Start with a model that, in some fashion,
reflects the truth about the random
phenomenon, and then develop a plan for
imitating – or simulating – a number of
repetitions of the procedure.
1.
Method 1 is slow, can be costly, and often
impractical or difficult.
 Method 2 requires that we know something
about the rules of probability and may not be
possible.
 Method 3 is usually quicker than repeating
the real procedure, especially when we can
use a calculator or computer. Simulation also
allows us to get reasonably accurate results
and answer questions that are more difficult
when done with formal mathematical
analysis.

 We
are interested in estimating the
likelihood that a couple will have a girl
among their first four children. Develop a
method for imitating births using
› A coin
› A six-sided die
› A deck of cards
› A Random Digit Table
 What
else could we use?
 Simulation
is an effective tool for finding
the likelihood of complex results once we
have developed a trustworthy model.
 We can use Random Digit Tables, graphing
calculators, computers or other items
(cards, coins, dice) to simulate many
repetitions quickly.
1.
2.
3.
4.
5.
State the problem or describe the
random phenomenon.
State the assumptions.
Assign digits to represent outcomes
(i.e. odd number is a head or a boy
and even number is a tail or a girl.
This depends on the problem you are
trying to simulate).
Simulate many repetitions (or trials).
State your conclusions.
 Events
are independent if the results
of one trial do not effect the results of
the next.
 For example, if you toss a coin and
get a head, that has nothing to do
with what you will get when you toss
the coin again.
 If you roll a die and get a 3, that has
nothing to do with what you will get
on the next roll.

We want to choose a person at random from
a group of which 70% are employed. How
would we assign numbers to individuals.
› 0, 1, 2, 3, 4, 5, 6 employed
› 7, 8, 9 unemployed
› We could assign any three digits to unemployed
as long as they are distinct
› We could also do the following:
 00,01,…, 69employed
 70, 71,…99unemployed
› This is less efficient because it uses twice as many
digits and 10 times as many numbers.
 We
want to choose a person at random
from a group of which 73% are employed.
How would we assign numbers to
individuals.
› 00, 01, …, 72 employed
› 73, 74, …, 99 unemployed
› We had to assign 73 of the 100 numbers to be
employed to get 73%. We could also do other
assignments, but doing it by groups is easiest.
 We
want to choose a person at random
from a group of which 50% are employed,
20% are unemployed, and 30% are not in
the labor force. How would we assign
numbers to individuals.
› 0, 1, 2, 3, 4employed
› 5, 6unemployed
› 7, 8, 9not in labor force
› The assignment of digits is not important. What
is important is how many digits are assigned to
each outcome.
 Orders
of frozen yogurt flavors (based on
sales) have the following relative
frequencies: 43% chocolate, 38% vanilla
and 19% strawberry. How could we
simulate customers entering the store and
buying yogurt?
› 00 to 42 would represent the outcome
chocolate.
› 43 to 80 would represent the outcome vanilla.
› 81 to 99 would represent the outcome
strawberry.
A
couple will have children until they have
4 children or a girl, whichever comes first.
 What is the likelihood that they would have
a girl?
 We will assume that each child is equally
likely to be a boy or a girl and that the sex
of each successive child is independent.
 How should we assign the digits?
› 0, 1, 2, 3, 4  girl or even  girl
› 5, 6, 7, 8, 9  boy or odd  boy

Let’s do the simulation letting girls be 0-4!
Start at line 130 in Table B and lets do 14
simulations.
690
51 64 81
7871
74
0
BBG
BG BG BG BBBG BG G
951
784
53 4 0 64
8987
BBG
BBG BG G G BG
BBBB
 So 13 of the 14 trials resulted in the birth of a
girl. Hence this strategy will produce a girl
13/14 or about 93% of the time. The actual
probability of having a girl would be 93.75%.
Using the randInt
Command!
 Let’s
try problems 6.11, 6.12, 6.13, 6.16, 6.20

Chance behavior is unpredictable in the short
run but has a regular and predictable pattern
in the long run.

Probability is the long-run proportion of
repetitions on which an event occurs.
 Count
Buffon in the 1700's
› Tossed a coin 4040 times and got 2048 heads
for a proportion of 0.5069.
 John
Kerrich imprisoned in WWII
› Tossed a coin 10,000 times and got 5067 heads
for a proportion of 0.5067
 Karl
Pearson around 1900
› Tossed a coin 24,000 times and got 12,012
heads for a proportion of 0.5005



We can never observe a probability exactly.
Mathematical probability is an idealization
based on imagining what would happen in a
long series of trials.
Remember the following:
› We must have a long series of independent trials,
i.e. the outcome of one trial must not influence the
outcome of another trial.
› The idea of probability is empirical, i.e. we must
observe many trials to estimate a probability.
› Computer simulation is very useful because it
allows us to see the results of many trials since it
often takes several hundred trials to determine the
probability of an outcome.
 What
is the sample space when we roll
two dice?
 Sometimes
the sample space varies
depending on what exactly you're asking
for!
 What is the sample space when we record
four coin tosses?
 What is the sample space when we toss a
coin and roll a die?
 What
is the sample space when we record
four coin tosses?
 What is the sample space when we toss a
coin and roll a die?
 Being
able to properly enumerate the
outcomes in a sample space is critical to
determining probabilities. Sometimes it's
helpful to use a tree diagram.

For example:
› How many ways can I toss a coin and roll a die?
› How many ways can I toss a coin and draw a
card from a standard 52-card deck?
› How many ways can I roll a die and draw a card
from a standard 52-card deck?

How do we describe probabilities
mathematically? Rather than try to give
“correct” probabilities by recording large
numbers of trials, we start by laying down
facts that must be true for any assignment of
probabilities.
Disjoint (mutually exclusive) events A and B
The Complement of Event A
Intersection of A and B (A⋂B)

Union – set of all outcomes in A or B
› Written A⋃B and read “A union B.”

Intersect or intersection – set of all outcomes that
are in A and B.
› Written A⋂B and read “A intersect B” or the intersection of A and B.”

Let’s do some examples.
› Let A = {0, 1, 2, 5, 7, 9}, B = {1, 3, 4, 5, 6, 8, 9} and
= {1, 3, 5, 7, 9}
 A⋃B =
 A⋂B =
 A ⋂C =
 B ⋃C =
 B ⋂C =
 A ⋃ C=
C

Union – set of all outcomes in A or B
› Written A⋃B and read “A union B.”

Intersect or intersection – set of all outcomes that
are in A and B.
› Written A⋂B and read “A intersect B” or the intersection of A and B.”

Let’s do some examples.
› Let A = {0, 1, 2, 5, 7, 9}, B = {1, 3, 4, 5, 6, 8, 9} and
= {1, 3, 5, 7, 9}
 A⋃B = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
 A⋂B = {1, 5, 9}
 A ⋂C = {1, 5, 7, 9}
 B ⋃C = {1, 3, 4, 5, 6, 7, 8, 9}
 B ⋂C = {1, 3, 5, 9}
 A ⋃ C= {0, 1, 2, 3, 5, 7, 9}
C

What is the probability for rolling a 5 with a
pair of dice?
 Find
the following probabilities.
› The first digit is one
› The first digit is odd
› The first digit is more than six
› Compare this last probability with the
probability that a random first digit is greater
than six
 Successive
coin tosses
 A single coin toss
 Taking an IQ test twice
 Dealing playing cards
 Spinning a number on a spinner
A rapid HIV test has probability about 0.004 of
producing a false-positive.
 If 200 hundred people are tested, what is the
probability that at least 1 false-positive will
occur?

› Based on what we are given, we know there is a 0.996
probability that we will not get a false positive for a single
person using the complement rule.
› There are too many ways that at least 1 person could receive
a false-positive. So we need to use the complement rule.
Hence:
P(at least 1 false-positive) = 1 – P(no false-positives out of 200)
=1-0.996200
=1-0.4486=0.5514

6.40, 6.41, 6.50, 6.51, 6.53, 6.55, 6.57, 6.59

Let A be the event that Joseph gets a hit, let B
be the event that Joseph gets a walk, let C
be the event that Joseph flies out, let D be
the event that Joseph grounds out, and let E
be the event that Joseph gets on base
because of an error. If
P(A) = 0.27, P(B) =
0.10, P(C) = 0.31, P(D) = 0.28, and P(E) = 0.04,
then find the following.
P(A or B or E) =
 P(C or D) =
 P(C or D or E) =


Let A be the event that Joseph gets a hit, let B
be the event that Joseph gets a walk, let C
be the event that Joseph flies out, let D be
the event that Joseph grounds out, and let E
be the event that Joseph gets on base
because of an error. If
P(A) = 0.27, P(B) =
0.10, P(C) = 0.31, P(D) = 0.28, and P(E) = 0.04,
then find the following.
P(A or B or E) = P(A) +P(B) + P(E) =0.27+0.10+0.04 = 0.41
 P(C or D) = P(C) + P(D) = 0.31+0.28 = 0.59
 P(C or D or E) = P(C) + P(D) + P(E) = 0.31+0.28+0.04 = 0.63


This works because the events are disjoint!



Suppose Event A has a 0.2 chance of occurring and Event B has a 0.4 chance of
occurring. If we add them, we would get that there is a 0.6 chance of A or B
occurring, but that is wrong. Why?
Because both events occur at the same time sometimes. That means that the
area where they overlap is counted twice and must be taken out!
So if we know that the Event that A and B both occur has a chance of 0.12, then
we know that the Chance of A or B occurring would be 0.2 + 0.4 – 0.12 = 0.48.
See the diagram!



Suppose Event A has a 0.2 chance of occurring and Event B has a 0.4 chance of
occurring. If we add them, we would get that there is a 0.6 chance of A or B
occurring, but that is wrong. Why?
Because both events occur at the same time sometimes. That means that the
area where they overlap is counted twice and must be taken out!
So if we know that the Event that A and B both occur has a chance of 0.12, then
we know that the Chance of A or B occurring would be 0.2 + 0.4 – 0.12 = 0.48.
See the diagram!
If A and B are disjoint, P(A and B) (also written as P(A⋂B)) has no
outcomes in it. Since it has no outcomes, it has a probability of 0. So
this rule works for all events and is a more generalized form of rule 3.
Let’s look at an example.
 Deborah estimates that she has a 0.7 chance of
being made a partner while Matthew has a 0.5
chance of being made a partner. She also estimates
that there is a 0.3 chance that they would both be
made partners. Answer the following:
 P(at least one is made a partner) =


P(neither is made a partner) =
Let’s look at an example.
 Deborah estimates that she has a 0.7 chance of
being made a partner while Matthew has a 0.5
chance of being made a partner. She also estimates
that there is a 0.3 chance that they would both be
made partners. Answer the following:
 P(at least one is made a partner) = P(D or M) =
= P(D) +P(M) –P(D ⋂M)
= 0.7 + 0.5 – 0.3 = 0.9


P(neither is made a partner) = 1 – P(D or M) =
= 1 – 0.9 = 0.1
 Fill
in what we know!
Matthew
Promoted
Deborah
Promoted
Not Promoted
Total
0.3
0.7
0.5
1
Not Promoted
Total
 Complete
the table!
Matthew
Deborah
Promoted
Not Promoted
Total
Promoted
0.3
0.4
0.7
Not Promoted
0.2
0.1
0.3
Total
0.5
0.5
1
NONE
20%
Coffee
20%
10%
20%
5%
5%
Tea
5%
Cola
15%
What’s the probability of drawing an Ace out
of a deck of cards?
 You have already drawn four cards and you
have a King, 2 Queens and a Jack in your
hand. Now what is the probability of drawing
an Ace out of the deck on the next draw?
 This is called a conditional probability
because it is finding the probability of one
thing occurring (drawing an Ace) given that
we know that another event (we already
have a King, 2 Queens and a Jack) has
already occurred.

The notation for a conditional probability is
P(A|B) and we read it “ the probability of A
given that B has already occurred.”
 Let’s look at an example on the next slide.


The table below breaks down student grades awarded
at a university by grade and the school that the
student took the class in. Let A = the grade comes
from an Engineering and Physical Sciences class, B =
the grade is below a B, C = the grade is an A, and D =
Liberal Arts class.
Find P(B) =
 Find P(B|A) =
 Find P(C) =
 Find P(C|D) =


The table below breaks down student grades awarded
at a university by grade and the school that the
student took the class in. Let A = the grade comes
from an Engineering and Physical Sciences class, B =
the grade is below a B, C = the grade is an A, and D =
Liberal Arts class.
Find P(B) = 3,656/10,000 = 0.3656
 Find P(B|A) = 800/1,600 = 0.5
 Find P(C) = 3,392/10,000 = 0.3392
 Find P(C|D) = 2,142/6,300 = 0.34

Since P(A⋂B) = P(B ⋂A), P(A⋂B) = P(B) P(A|B).
 This rule is true because in order for both A
and B to occur, one of them must occur, then
we have to multiply by the probability that
the other occurs given that the first already
occurred.


Downloading Music
› 29% of internet users download music files and 67% of
downloaders say they don’t care if the music is
copyrighted. So the percent of internet users who
download music (event A) and don’t care about
copyright (event B) is 67% of the 29% who download or
(0.67)(0.29) = 0.1943 = 19.43%. The multiplication rule
expresses this as P(A and B) = P(A) P(B|A).

Slim wants two diamonds
› Slim needs to draw two diamonds in a row. Slim can
see 11 cards and 4 of these are diamonds. What’s the
probability that Slim can draw two diamonds in a row?
› P(drawing two diamonds) = P(drawing one diamond) P(drawing another diamond given he
already drew one)
› So P(drawing two diamonds) = (9/41)(8/40) = 72/1640 = 0.044
 We
can rearrange the general
multiplication rule [ P(A and B) = P(A)
P(B|A) ]to get a formula for conditional
probability.
 We
can also get
Let A = grade is an A and B = a liberal arts
course.
 What’s the conditional probability that a
grade is an A, given that it comes from a
liberal arts course?
 P(A|B)= P(A and B) / P(B) = (2,142/10,000) /
(6,300/10,000) = 0.34

Let A = grade is an A and B = not a liberal arts
course.
 What’s the conditional probability that a
grade is an A, given that it does not come
from a liberal arts course?
 P(A|B)= P(A and B) / P(B) = (1,250/10,000) /
(3,700/10,000) = 0.3378




P(A and B and C) = P(A) P(B|A) P(C|A and B)
P(A and B and C and D) = P(A) P(B|A) P(C|A and B) P(D|A
and B and C)
And so on…

Online chat rooms are dominated by the
young. If we look at adult Internet users only,
47% of the 18 to 29 age group chat, as do
21% of those aged 30 to 49 and just 7% of
those 50 and over. Total 29% of adult Internet
users are aged 18 to 29, another 47% are 30
to 49, and the remaining 24% are 50 or over.

The probability of reaching the end of any
complete branch is the product of the
probabilities written on its segments.
What
percent of adult chat room participants
are aged 18 to 29?

What is the probability that a randomly
chosen user of the Internet participates in
chat rooms?
0.2518 is
the
probability
because
we would
add these
three up!

6.80, 6.83, 6.86, 6.87, 6.88, 6.90, 6.91