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Chapter 5
5.2 Probability
Warm up
Suppose I want to choose a simple random sample of
size 6 from a group of 60 seniors and 30 juniors. To do
this, I write each person’s name on an equally-sized
piece of paper and mix the papers in a large grocery bag.
Just as I am about to select the first name, a thoughtful
student suggests that I should stratify by class. I agree,
and we decide it would be appropriate to select 4
seniors and 2 juniors. However, because I have already
mixed up the names, I don’t want to have to separate
them all again. Instead, I will select names one at a time
from the bag until I get 4 seniors and 2 juniors. Design
and carry out a simulation using Table D to estimate the
probability that you must draw 10 or more names to get
4 seniors and 2 juniors.
State: What is the probability that it takes 10 or
more selections to get 4 seniors and 2 juniors?
Plan: Using pairs of digits from Table D, we’ll label
the 60 seniors 01–60 and the 30 juniors 61–90.
Numbers 00 and 91–99 will be skipped. Moving left
to right across a row, we’ll look at pairs of digits until
we have 4 different labels from 01–60 and 2
different labels from 61–90. Then we will count how
many different labels from 01–90 we looked at.
Do: Here is an example of one repetition, using line 101 from Table D:
19 (senior) 22 (senior) 39 (senior) 50 (senior) 34 (senior) 05 (senior) 75
Dot Plot
(junior) 62 (junior)
Collection 2
In this trial, it took exactly 8 selections to get at least 4 seniors and at
least 2 juniors. Here are the results of 50 trials:
6
7
8
9 10 11 12 13 14
NumSel
Conclude: In the simulation, 11 of the 50 trials required 10 or more
selections to get 4 seniors and 2 juniors, so the probability that it takes
10 or more selections is approximately 0.22.
5.2 Probability Models
• Sample space: set of all possible outcomes
• Probability model: a mathematical description of a
random phenomenon consisting of two parts: a sample
space and a way of assigning probabilities
• What is the sample space of rolling two die?
– Display the outcomes
– How many outcomes?
– What are the possible sums of the die?
Example: Roll the Dice
Give a probability model for the chance process of rolling two fair, sixsided dice – one that’s red and one that’s green.
Sample
Space
36 Outcomes
Since the dice are fair, each
outcome is equally likely.
Each outcome has probability
1/36.
Probability models allow us to find the
probability of any collection of outcomes.
Definition:
An event is any collection of outcomes from some chance process.
That is, an event is a subset of the sample space. Events are usually
designated by capital letters, like A, B, C, and so on.
If A is any event, we write its probability as P(A).
In the dice-rolling example, suppose we define event A as “sum is 5.”
There are 4 outcomes that result in a sum of 5.
Since each outcome has probability 1/36, P(A) = 4/36 = 1/9.
Suppose event B is defined as “sum is not 5.” What is P(B)?
P(B) = 1 – 4/36
= 32/36 = 8/9
Basic Rules of Probability
• The probability of any event is a number between 0
and 1.
0 ≤ P(A) ≤ 1.
• All possible outcomes together must have probabilities
whose sum is exactly 1.
P(S) = 1.
• If all outcomes in the sample space are equally likely,
the probability that event A occurs can be found using
the formula
number of outcomes corresponding to event A
P(A) =
total number of outcomes in sample space
Complement Rule P(Ac) = 1- P(A)
• The probability an event does not occur is 1
minus the probability it does occur
• Ex: If the probability of you getting a 5 on the
AP exam is 12%, what is the probability that
you do not get a 5?
Basic Rules of Probability
Two events A and B are mutually exclusive (disjoint) if
they have no outcomes in common and so can never
occur together—that is, if P(A and B ) = 0.
• If two events have no outcomes in common, the probability that
one or the other occurs is the sum of their individual probabilities.
• Addition rule for mutually exclusive events: If A and B are
mutually exclusive,
P(A or B) = P(A) + P(B).
Mutually exclusive (Disjoint) events
• Getting an A and C in statistics the same 6
weeks.
• Rolling a sum of 7 or 11
• Marital status of women ( divorced, single,
married, widowed)
• Suppose the probability that a randomly
selected students is a sophomore is.20 and
the probability that he or she is a junior is .3.
What is the probability that the student is
either a sophomore or a junior?
Randomly select a student who took the 2013 AP® Statistics
exam and record the student’s score. Here is the probability
model:
Score:
1
Probability: 0.235
2
0.188
3
0.249
4
0.202
5
0.126
Problem:
(a) Show that this is a legitimate probability model.
(a) All the probabilities are between 0 and 1 and the sum of
the probabilities is 1, so this is a legitimate probability model.
(b) Find the probability that the chosen student scored 3 or
better. By the addition rule, P(3 or better) = 0.249 + 0.202 + 0.126 =
0.577.
By the complement rule and addition rule, P(3 or better) = 1 –
P(2 or less) = 1 – (0.235 + 0.188) = 1 – 0.423 = 0.577.
Warm up
1. Mrs. R’s TV can pick up 120 stations. At 8pm, 42
stations are showing commercials, 18 sports, 25
movies, 7 news shows, 13 comedy programs, 12
drama programs, and 3 are sales programs (like
QVC). A station is chosen at random. Find the
following probabilities:
a) Choosing a news show 7/120
b) Choosing a commercial 42/120 = 7/20
c) A comedy or drama 13/120 + 12/120 = 25/120 = 5/24
d) Sports, movie or news 18/120 + 25/120 +7/120 = 50/120 = 5/12
e) Not a sales program 1 – 3/120 = 117/120 = 39/40
f) Neither sports nor a movie 1 – (18/120 + 25/120) = 77/120
2. Choose an American adult at random. Define two events:
A = the person has a cholesterol level of 240 mg/dl of blood or above
(high cholesterol)
B = the person has a cholesterol level of 200 to 239 mg/dl (borderline
high cholesterol)
According to the American Heart Association, P(A) = 0.16 and P(B) =
0.29
a) Explain why events A and B are mutually exclusive.
A person cannot have a cholesterol level of both 240 or above and between 200
and 239 at the same time.
b) Say in plain language what the event “A or B” is. What is P(A or B)?
A person has either a cholesterol level of 240 or above or they have a
cholesterol level between 200 and 239. P(A or B) = 0.16 + 0.29 = 0.45
c) If C is the vent that the person chosen has normal cholesterol
(below 200 mg/dl), what’s P(C)?
P(C) = 1 – 0.45 = 0.55
Two-Way Tables and Probability
When finding probabilities involving two events, a two-way table can
display the sample space in a way that makes probability calculations
easier.
Consider the example on page 309. Suppose we choose a student at
random. Find the probability that the student (a) has pierced ears.
(b) is a male with pierced ears.
(c) is a male or has pierced
ears.
Define events A: is male and B: has pierced ears.
(a)
(b) Each
(c)
We want
student
to find
is equally
P(male likely
or
and
pierced
pierced
to be ears),
chosen.
ears),
that
that
103
is,is,
students
P(A
P(A
orand
B).have
There
B).
pierced
are
ears.
in
So,
theP(pierced
classofand
ears)
103
=
individuals
P(B)
103/178.
with
pierced
ears.There
Look90atmales
the intersection
the
“Male”
row=and
“Yes”
column.
are 19 males
with pierced
ears. So,
P(A
and B)
= 19/178.
However,
19 males
have pierced
ears
– don’t
count
them twice!
P(A or B) = (19 + 71 + 84)/178. So, P(A or B) = 174/178
Who owns a home?
What is the relationship between educational achievement and home
ownership? A random sample of 500 U.S. adults was selected. Each
member of the sample was identified as a high school graduate (or not)
and as a home owner (or not). The two-way table displays the data.
Homeowner
Not a homeowner
Total
High school
graduate
221
89
310
Not a high school
graduate
119
71
190
Total
340
160
500
Suppose we choose a member of the sample at random. Find the probability that the
member:
(a) is a high school graduate. Because 310 of the 500 members of the sample
graduated from high school, P(G) = 310/500.
(b) is a high school graduate and owns a home.
Because 221 of the 500 members of the sample graduated from high
school and own a home, P(G and H) = 221/500.
(c) is a high school graduate or owns a home.
Because there are 221 + 89 + 119 = 429 people who graduated from high school or own a
home, P(G or H) is 429/500. Note that it is inappropriate to compute P(G) + P(H) to find this
probability because the events G and H are not mutually exclusive—221 people both are high
school graduates and own a home. If you did add these probabilities, the result would be
650/500, which is clearly wrong because it results in a probability that is greater than 1.
General Addition Rule for Two Events
We can’t use the addition rule for mutually exclusive events unless the
events have no outcomes in common.
General Addition Rule for Two Events
If A and B are any two events resulting from some chance
process, then
P(A or B) = P(A) + P(B) – P(A and B)
Who owns a home? Using the previous example,
use the general addition rule to find the probability
that a random person is a homeowner or graduate.
P(G or H) = P(G) + P(H) – P(G and H)
= 310/500 + 340/500 – 221/500 =
429/500.
This matches our previous result.
Venn Diagrams and Probability
Because Venn diagrams have uses in other branches of mathematics,
some standard vocabulary and notation have been developed.
The complement AC contains exactly the outcomes that are not in A.
The events A and B are mutually exclusive (disjoint) because they do not
overlap. That is, they have no outcomes in common.
Venn Diagrams and Probability
The intersection of events A and B (A ∩ B) is the set of all outcomes
in both events A and B.
The union of events A and B (A ∪ B) is the set of all outcomes in either
event A or B.
Hint: To keep the symbols straight, remember ∪ for union and ∩ for intersection.
Use the same two-way table from before summarizing the
relationship between educational status and home ownership from
the previous example. Our events of interest were G: is a high school
graduate and H: is a homeowner. Make a Venn Diagram.
The four distinct regions in the Venn diagram below correspond to
the four cells in the two-way table as follows:
Region in Venn
diagram
In the intersection
of two circles
Inside circle G,
outside circle H
Inside circle H,
outside circle G
Outside both circles
In words
In symbols
Count
Graduate and owns home
G ∩ H
221
Graduate and doesn’t own
home
Not a graduate but owns
home
Not a graduate and doesn’t
own home
G ∩ HC
89
GC ∩ H
119
GC ∩ HC
71
Phone usage
According to the National Center for Health Statistics
(http://www.cdc.gov/nchs/data/nhis/earlyrelease/wireless201306.pdf), in
December 2012, 60% of U.S. households had a traditional landline telephone,
89% of households had cell phones, and 51% had both. Suppose we randomly
selected a household in December 2012.
Problem:
(a) Make a two-way table that displays the sample space of this chance
process.
(b) Construct a Venn diagram to represent the outcomes of this chance
process.
(c) Find the probability that the household has at least one of the two types of
phones.
(d) Find the probability that the household has a cell phone only.
a) We will define events L: has a landline and C: has a cell phone.
Cell phone
0.51
0.38
0.89
Landline
No landline
Total
No cell phone
0.09
0.02
0.11
Total
0.60
0.40
1.00
C
L
.09
0.51
0.38
0.02
(c) To find the probability that the household has at least one of the two types of
phones, we need to find the probability that the household has a landline, a cell phone, or
both.
P(L ᴜ C) = P(L) + P(C) - P(L ∩ C) = 0.60 + 0.89 – 0.51 = 0.98
There is a 0.98 probability that the household has at least one of the two types of
phones.
(d) P(cell phone only) = P(LC ∩ C) = 0.38. There is a 0.38 probability that the household
is cell-phone only.