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CHAPTER 6 PROBABILITY
How can the likelihood of an event be measured?
CHANCE EXPERIMENTS AND EVENTS 6-1

What is a chance experiment and how can
those chances be represented?

Chance Experiment—any activity where the
outcome has two or more possibilities

Sample Space—the set of all possible
outcomes

Tree Diagram—one method of finding the
sample space that shows/details all the
outcomes
EXAMPLE

Show all possible outcome for the situation:
a woman and a man walk into a sport
equipment store. Each one could buy a tennis
racket, a soccer ball or a Frisbee.
Tennis racket
Soccer ball
man
woman
Frisbee
Tennis racket
Soccer ball
Frisbee

Set notation—
 (event1,
event2, . . . )

Event—any collection of outcomes from the
sample space of a chance experiment

Simple event—consists of exactly one outcome
from the sample space

The complement of an event A—all the outcomes that
are not in A


Noted as A’, Ac , or A
A or B—something which is in the sample space of A or
the sample space of B

AUB
(A union B)

A and B—something which is in the intersection of the
sample space of A and of B
 A  B (A intersect B)

Disjoint or mutually exclusive events—two events that
have no common outcomes
VENN DIAGRAMS
Shaded area represents the indicated statement
A or B
A and B
Disjoint Events
A’
Homework Pg 233-235 3, 4, 7, 10, 12
THE DEFINITION OF PROBABILITY
What is probability?
6-2
Probability
# of favorable outcomes
P( E ) 
# of possible outcomes
Called the classical approach and works well
when you have a finite number of outcomes

Law of Large Numbers (James Bernoulli)

As the number of repetitions of a chance
experiment increases the chance that the
relative frequency of an event will differ from
the true probability of the event by more than a
small number approaches zero
In English:
The more trials you do for an experiment, the closer the rel. freq. you get
will be to the true probability.

Relative Frequency Approach
# of times E occurs
P( E ) 
# of trials

Subjective Approach
Personally assigned values which are not replicable
 not as much value is placed on these
BASIC PROPERTIES OF PROBABILITY

6-3
What are the basic properties of probability?
1)
For any event E
0≤p(E)≤1
2)
If S is the sample space for an experiment
then P(S) = 1
3)
If two events are disjoint, P(E or F) = P(E) +P(F)
4)
For any event E
P(E) + P(not E) = 1
 P(E) =1 – P(not E) or P(not E) = 1- P(E)
Homework P 248-251 evens
CONDITIONAL PROBABILITY
6-4
What is conditional probability and how can it be
found?
Conditional Probability occurs when the likelihood
that one event will occur is changed by what
has happened previously.
EXAMPLE
There is a disease that effects .1% of the population. A test to
determine if you have the disease exists. The test can give a
false positive 20% of the time.
IF
E= the individual has the disease
F= the individual tested positive for the disease
What is
P(E)=.001=.1%
P(E|F)=.8=80%
The probability of E given that F occurred
The occurrence of E is unlikely in the general population however knowing that
the test is positive it is very likely that you have the disease.
Conditional Probability Formula:
P( E and F) P( E  F )
P( E | F ) 

P(F)
P( F )
Read the probability of E given F
Hint:
intersection = multiplication
union = addition

Example
Consider the population to be all families with two children. Assuming that
there is an equal chance of having a boy or a girl there are 4 possible
outcomes. GG, BB, GB, BG (ordered by age)
1) What is the probability of choosing a family with two girls given the family
has at least one girl?
2) What is the probability that they have 2 girls given that the oldest is a girl?
EXAMPLE
In 1912, the titanic sank. Almost 1500 people died, most of
them men. Was that because a man was less likely to survive
than a woman or because men outnumbered women 3 to 1?
Female Survival Rate p(S|F)
Titanic Survival Data:
Male
Female
Totals
Survived
367
344
711
Died
1364
126
1490
totals
1731
470
2201
surviving female 344

 73%
female
470
Male Survival Rate
P(S|M)
surviving male 367

 21%
male
1731
Overall Survival Rate P(S|on the ship)
survived
711

 32%
on the ship 2201
Lore is likely to be correct. Women and children first!
Homework pg. 258-260 even
INDEPENDENCE
6-5
What are independent events and how are their
probabilities found?
Independent events have no bearing on each
other even if you know the outcome of the 1st
event
P(E|F) = P(E)
this also implies
P(not E|F) = P(not E)
P(E| not F) = P(E)
P(not E|not F) = P(not E)
Basically logical
Probability of 2 independent events occurring
follows from:
P( E  F )
P( E | F ) 
P( F )
Cross multiply:
P( E  F )  P( E | F ) P( F )
But if E and F are Independent P(E|F) = P(E) we substitute to find
P( E  F )  P( E ) P( F )
Replacement—not removing a selection from the
population (look at it and put it back).
What is the probability of getting a red marble on the second try
given that you had 5 red and 5green marbles in the bag and
replaced the first marble after choosing it.
Without Replacement—removing the selection
from the population once it is selected.
What is the probability of getting a red marble on the second try
given that you had 5 red and 5green marbles in the bag and did
not replace the first marble after choosing it.
Homework Pg 267-270 36 – 52 even and
Pg 259 #33
SOME GENERAL PROBABILITY RULES
6-6
What are some basic or general probability rules?
THE UNION OF EVENTS:
P(A U B) = P(A) + P(B) – P(A ∩ B)
Inclusive—events that share some solutions
0≤P(A∩B)≤1
Mutually exclusive events do not have any overlap
P(A∩B)=0
REMEMBER:
P(E∩F)=P(E|F)P(F) if they are independent
but =P(E)P(F)
Law of Total Probability
if B1, B2, …, Bn are disjoint events with
P(E∩B1) + P(E∩B2) + … + P(E∩Bn) = 1
then
P(E) = 1
P(E) = P(E|B1)P(B1) +P(E|B2)P(B2) + … + P(E|Bn)P(Bn)
Baye’s Rule
if the items are disjoint and you want the probability of P(B1|E) but have P(E|B1)
P( B1  E ) P( E  B1 )

P( E )
P( E )
Knowing
P(B1 | E ) 
And
P( E  B1 )
P(E | B1 ) 
P( E )
Or
P(B1 )P(E | B1 )  P( E  B1 )
Then by substitution:
P(B1|E) =
P(B1)P(E|B1)
P(E|B1)P(B1) +P(E|B2)P(B2) + … + P(E|Bn)P(Bn)
Example pg 279 #6.27
Two shipping services offer overnight delivery of parcels, and both promise delivery before 10AM. A
mail-order catalog company ships 30% of its overnight packages using shipping service 1 and 70%
using service 2. Service 1 fails to meet the 10AM delivery promise 10% of the time, whereas
service 2 fails to deliver by 10 AM 8% of the time. Suppose you made a purchase from this
company and were expecting your package by 10 AM but it is late. Which service is more likely to
have been used—ie who should you call to see where your important package is?
S1 = service 1
S2 = service 2
L = late
We know P(S1) = .3 P(S2) = .7
P(L|S1) = .1 P(L|S2) = .08
Since you know the package is late you need to find P(S1|L) and P(S2|L)
P( S1 | L) 
P( L | S1 ) P( S1 )
(.1)(. 3)
.03


 .3488
P( L | S1 ) P( S1 )  P( L | S 2 ) P( S 2 ) (.1)(. 3)  (.08)(. 7) .086
P ( S 2 | L) 
P( L | S 2 ) P( S 2 )
(.08)(.7)
.056


 .6512
P( L | S1 ) P( S1 )  P( L | S 2 ) P( S 2 ) (.1)(.3)  (.08)(.7) .086
So you should call service two to see where the package is because even though they have a lower
overall percentage rate (.08), they have 70% of the company’s business so 65% of the late
packages from this company are from service 2.
(it’s a lot of math this time, but if you are constantly having to call it would be beneficial to know
who to call first and save time overall by only making 1 call 65% of the time)
Homework pg 280-282 even
ESTIMATING PROBABILITIES EMPIRICALLY AND
USING SIMULATIONS
6-7
How do you accurately estimate probabilities?
Simulation provides a method of estimating
probabilities that we do not have the time or
resources to determine analytically and are not
practical to observe or estimate empirically.
Simulation generates “observations” by
performing an experiment that is as similar as
possible in structure to the real situation.
EXAMPLE:
A professor wants to simulate the possible scores on a 20 question
true/false quiz when the students are guessing. Rather than finding 500
students to write down 20 answers and then scoring each paper.
You could place 20 balls 10 red/10 blue in a box, allow blue to represent
true, make a selection, replace the ball and repeat 20 times “per quiz”,
record the # of corrects. Repeat 1000 times and construct a table of
probability.
Why 20 balls with 10 of each?
to make it random but equally likely to pick either color
OR
You could use a random number table allowing evens to be correct and
odds incorrect break the table into groups of 20 and count the number of
corrects in each.
Homework pg 291 – 294 even
Review Pg. 295 – 297 even and 81