Chapter 5 (Probability)
Summary
David Cheuvront, Ruben Andrade, Noel Casino
What is Probability?

Probability is the chance that a certain outcome of a random process
that describes how many times it will happen over the course of many
trials.

We can use probability because of the large law of numbers, which
states that the proportion of times that a particular outcome occurs in
many repetitions will approach a single number.

Probability can be expressed in a few ways: ½, .50, and 50%. All these
values are the same.

An example of probability is the chance of getting heads when you toss a
fair coin. There is a .50 or 50% chance that you will get heads.

Remember that probability is a long-run phenomena. Regularity only
emerges after many repetitions.
Probability Problem
1.
Suppose the probability of bouncing a ping-pong ball into
a red solo cup is 0.4.
A.
What does this probability mean?
Answer: The probability in this situation means that over the
course of many trials, the probability that the ping-pong ball
will land in the cup is 0.4 or 40%.
B.
Question B. Because the probability is 0.4, am I
guaranteed to make 4 out of 10 shots?
Answer: No. The amount of times you make it into the cup
could be greater or less than 4 because of random chance.
Basic Rules of Probability

The Probability of any event is a number between 0 and 1

All possible outcomes together must have probabilities that add up to 1.

If all outcomes in the sample space are equally likely, the probability
that event A occurs can be found using the formula: Outcomes with
Event A / Total Number of Outcomes.

The probability that an event does not occur is 1 minus the probability
that the event does occur.

If two events have no outcomes in common, the probability that one or
the other occurs is the sum of their individual probabilities.
Basic Rules of Probability Problem
2.
Determine if the probability given to the following situations are
legitimate, that is if it satisfies the rules of probability.
A.
Pick a random person and record their gender and employment
status, P(Woman, Employed) = 0.40, P(Woman, Employed) = 0.10,
P(Man, Unemployed) = 0.05, P(Man, Employed) = 0.50.
Answer: They are not legitimate because 0.40+0.10+0.05+0.50 =
1.05, which
does not add up to zero
B.
Pick a random card from a deck. Record the suit the card is in.
P(Spade) = 13/52, P(Heart) = 12/52, P(Club) = 14/52, P(Diamond) =
13/52.
Answer: (13/52) + (12/52) + (14/52) + (13/52) = 52/52 = 1
1.
This situation is legitimate because the probabilities all add up to
Simulations

A simulation is an imitation of chance behavior. To
perform one, follow these 4 steps:
1.
State: Ask question about the chance process.
2.
Plan: Describe how to use chance to imitate one
repetition of the process. Decide what you want to
record at the end of each repetition.
3.
Do: Perform many trials of the simulation.
4.
Conclude: Use results to answer the question of
interest you created in step 1.
Simulation Problem
3.

A.
Suppose the chance of Ruben catching an iguana when he throws a net is 0.25.
How could we simulate his chance of catching an iguana with:
A deck of Cards
Answer: Let hearts represent him catching an iguana, and clubs and spades represent
him not catching one. Draw a card.
B.
Table D - line 112
Answer: Go to table D – line 112. Let numbers 00-24 represent him catching an iguana,
and 25-99 represent him not catching one. Read a two-digit number from Table D line
112.
C.
Using a random number generator.
Answer: Go to a random number generator and set the upper to 4 and the lower to 1.
Let the number 1 represent that he catches an iguana and numbers 2-4 represent that
he doesn’t.
Probability Models

Suppose we flip a fair coin 3 times. Let X=The amount of times you get a head
create the sample space and probability distribution for X.
Sample Space:
A sample space is the
set of all possible
outcomes.
The sample space for
this problem is:
HHH
HTH
HHT
HTT
TTT
THT
TTH
THH
Probability Model:
The probability Model is a description of some
chance process that consists of the sample space
and the probability for each outcome.
The Probability Model for this problem is:
HHH
HTH
HHT
HTT
TTT
THT
TTH
THH
Each outcome has a probability of 1/8 of
happening. The probability of all
outcomes should add up to 1.
Probability Models Cont.
We can graph the probability of
X = number of heads you get in 3
Coin Flips.
P(X=0) –
P(X=1) –
P(X=2) –
P(X=3) –
1/8
3/8
3/8
1/8
An event is any collection of outcomes
from the same chance process. This is
a subset of the sample space. Events
are usually designated by capital
letters, such as A,B,C, etc.
Define event A = “2 heads”
There are 3 outcomes in event A: HTH,
HHT, THH
Therefore, the probability of Event A is
3/8 or 0.375.
Mutual Exclusivity and the
Compliment Rule

Two events are Mutually Exclusive, or disjoint, if
they can never occur together. I.E. P(A and B) = 0

The Complement Rule is P(AC) = 1- p(A). Because we
know that events of A occurring and not occurring
always add up to 1, we know that the probability of
A not happening is 1 – the probability of A occurring.
We can use this rule to find out the complement of
A, which is the probability of A not occurring.
Mutually Exclusive Problem
4.
We sampled 400 truck owners at a chili cook-off sponsored by Chevy and
Ford. Only trucks made by Chevy or Ford were let in to the event. In the
following table, what event is mutually exclusive?
Truck Brand
4 Wheel Drive
2 Wheel Drive
Chevy
186
59
Ford
0
155
Answer: Having a truck made by Ford and having a 4 wheel drive is mutually
exclusive because the probability of having a four wheel drive truck made by ford
is 0. This could be a result of Ford not making a truck with a 4 wheel drive.
Complement Rule Problem
5.
The probability of a randomly sampled American being obese is 0.40. What
is the probability of a randomly sampled American NOT being obese?
Answer: By using the complement rule, we can find the answer to this question.
We know that the complement rule is P(AC) = 1- p(A). To answer this problem, we
need to find P(AC).
P(AC) = 1 – 0.40
P(AC) = 0.60

The answer to this question is that the probability of a randomly sampled
American NOT being obese is 0.60.
Two-Way Tables


Two-Way tables help display the probability two events happening.
Married
Male
Female
Total
Yes
218
316
534
No
180
86
266
Total
398
402
800
In this two way table, we display the number of people that are both married
and their genders. This helps display two variables and their outcomes easily.
Completing a Two-Way Table Problem
6.
Complete the Two-way Table shown below.
Type of Housing
Male
Female
Total
House
249
???
448
Apartment
92
160
252
Total
341
359
700
Answer: To complete the two way table, we must find the amount of women
living in a house. To do this, we subtract the total number of people living in a
house and the number of men living in a house.
249 + ? = 448
448 – 249 = 199

The number of females living in a house is 199. Using the information given to
us in this incomplete table, we can complete it.
“And” Notation

If we want to find the probability of something having one
characteristic AND another characteristic, we find the probability
of (A and B). The notation for this is P(A Ω B). This means that the
outcome has to meet both the requirements of A and B.

We will use the two-way table from question 6 to demonstrate
what we are looking for. Say we want to find the probability of a
person that is married AND a male, or P(Married Ω Male). We
would try to find the probability of randomly selecting the
Married
Male
Female
Total
highlighted cell.
Yes
218
316
534
No
180
86
266
Total
398
402
800
“And” Notation Continued

To do this, we would put the number of people that fit the requirements of
being married AND being a male, and divide it by the total amount of people.
218/800 = 0.2725
P(Married Ω Male) = 0.2725 or 27.25%
“And” Notation Problem
7.
Use the table below to find P(Dog Ω Male).
Animal
Male
Female
Total
Dog
572
228
800
Cat
97
103
200
Total
669
331
1000
Answer: To find this probability, we have to find the number of animals that are
both a dog AND a male and divide it by the total number of animals.
P(Dog Ω Male) = 572/1000
P(Dog Ω Male) = 0.572 or 57.2%
“Or” Notation

If we want to find the probability of something of having
one characteristic OR another characteristic, we find the
probability of (A or B). The notation for this is P(A U B).
We use this whenever we want to find out when an
outcome meets the requirements of A or B.

We will use the two-way table from question 6 again to
demonstrate what we are looking for. Say we want to find
the probability of a person that is married OR a male, or
P(Married U Male). We would try to find the probability of
randomly selecting the highlighted cells.
Married
Male
Female
Total
Yes
218
316
534
No
180
86
266
Total
398
402
800
“Or” Notation Continued

To find the probability, we would add up the number of
people that fit the requirement of being married OR being
male and then divide it by the total number of people
(218+316+180)/800 = 0.8925
P(Married U Male) = 0.8925 or 89.25%
*Be careful not to include the cell highlighted in purple twice! If you did, your
probability would be over 1, which, as previously stated, is impossible!*
“Or” Notation Problem
Use the following table to find P(Deer U Male).
Animal
Male
Female
Total
Deer
324
348
672
Moose
142
186
328
Total
466
534
1000
8.
Answer: To find the probability, we must add up the number of all the animals
that are either a deer OR a male and divide it by the total number of animals.
P(Deer U Male) = (324 + 142 + 348) / 1000
P(Deer U Male) = 0.814 or 81.4%
*Remember not to count the same value twice in your calculations!)*
Venn Diagrams

Venn Diagrams are very similar to two-way tables, as
they show the probability of two variables.
Conditional Probability

Conditional Probability is the probability that one event
happens given that another event is already known to
have happened. Lets say that we know that event B
already happened. The probability that event B happens
given that event A has happened is denoted by P(A І B)

To find the conditional probability of an event, use the
formula:
P(A І B) = P(A Ω B) / P(B)
Conditional Probability Problem
9.
Use the following table to find P(Male І Home).
Home?
Male
Female
Total
Home
44
45
89
Homeless
8
3
11
Total
52
48
100
Answer: We are trying to find the probability of a person being a male, given that
they have a home. To do this, we must find the P(Male Ω Home) and divide it by
P(Home).
P(Male І Home) = (44/100) / (89/100)
P(Male І Home) = 0.4944 or 49.44%
The General Multiplication Rule
 The
General Multiplication Rule is used to
find the probability that events A and B both
occur.
 The
formula for the Rule is: P(A Ω B) = P(A) x
P(A І B)
 This
rule is useful when a chance process
involves a sequence of outcomes. In these
cases, we can use tree diagrams to display the
sample space.
General Multiplication Rule Problem

According to our research, the probability that Noel is
from Vietnam is 0.64. The probability of a randomly
selected person from Vietnam being Chinese is 0.24. What
is the chance that Noel is from Vietnam and Chinese?

We know that P(Vietnam) = 0.64 and P(Vietnam І Chinese)
= 0.24

According to our general multiplication rule, P(Vietnam Ω
Chinese) = P(Vietnam) x P(Vietnam І Chinese).
P(Vietnam Ω Chinese) = 0.64 x 0.24
P(Vietnam Ω Chinese) = 0.1536 or 15.36%
Tree Diagrams

Tree diagrams rely heavily on the multiplication rule. They show all the
probabilities of each route.
Independent Events
 Two
events are independent if the occurrence of
one event does not change the probability that
the other event will happen. Events A and B are
independent if P(A І B) = P(A) x P(B І A) = P(B)
 The
multiplication rule for independent events is:
P(A Ω B) = P(A) x P(B)
Question 1!
1.
P(X) = 0.25 and P(Y) = 0.40. If P(X I Y) = 0.20, what is
P(Y I X)?
A.
0.10
B.
0.125
C.
0.32
D.
0.45
E.
0.50
Question 1 Answer
 The

answer is C: 0.32
P(X Ω Y) = P(X I Y) x P(Y) = 0.20 x 0.40 = 0.08
Then P(Y I X) = P(X Ω Y) / P(X) = 0.08 / 0.25 = 0.32
Question 2!
2.
Given the probabilities P(A) = 0.3 and P(A U B) =
0.7, what is the probability P(B) if A and B are
mutually exclusive? If A and B are independent?
A.
0.4, 0.3
B.
0.4, 4/7
C.
4/7, 0.4
D.
0.7, 4/7
E.
0.7, 0.3
Question 2 Answer
 The
answer is B: 0.4, 4/7
 If
A and B are mutually exclusive, P(A Ω B) = 0.
Thus, 0.7 = 0.3 + P(B) – 0, and so P(B) = 0.4.
 If
A and B are independent, then P(A Ω B) =
P(A) x P(B). Thus 0.7 = 0.3 + P(B) -0.3P(B), and
so P(B) = 4/7
Question 3
3.
You win a game if you flip a coin with heads
coming up exactly 50% of the tosses, Would you
rather flip 10 times or 100 times?
A.
10 times because 0.246 > 0.080
B.
10 times because of the Central Limit Theorem
C.
100 times because of the Law of Large Numbers.
D.
Your chance is the same.
Question 3 Answer

The answer is A: 10 times because 0.246 > 0.080

Binomcdf (10, 0.5, 5) = 0.246 while
Binomcdf(100, 0.5, 50) = 0.080. The probability
that you get exactly 50% of the coin tosses being
heads on throwing just 10 coins is higher than if
you were to do it 100 times.
Question 4!
4.
Given two events, E and F, such that p(E) = 0.340, P(F) = 0.450, and P(E U F)
= 0.637, then the two events are:
A.
Independent and mutually exclusive.
B.
Independent, but not mutually exclusive.
C.
Mutually Exclusive, but not independent.
D.
Neither independent nor mutually exclusive.
E.
There is not enough information to determine the answer.
Question 4 Answer

The answer is B:Independent, but not mutually
exclusive.

P(E U F) = P(E) = P(F) – P(E Ω F), so 0.637 = 0.340
+ 0.450 – P(E Ω F) and P(E Ω F) = 0.153. Since P(E
Ω F) does not equal 0, E and F are not mutually
exclusive.

P(E Ω F) = 0.153 = 0.340 x 0.450 = P(E) x P(F),
which implies that E and F are independent.
Question 5!
5.
There are two games involving flipping a fair coin. In the first
game, you win a prize if you can throw between 45% and 55%
heads; in the second game, you win if you can throw more than
60% heads. For each game, would you rather flip the coin 30
times or 300 times?
A.
30 times for each game
B.
300 times for each game
C.
300 times for the first game, 30 times for the second game
D.
The outcome of the games do not depend on the number of flips.
Question 5 Answer

The Answer is C: 300 times for the first game, 30
times for the second game

By the Law of Large Numbers, the more times you
flip a coin, the closer the relative frequency
tends to be to the probability of 0.50. With fewer
tosses, there is more chance of wide swings in the
relative frequency.
Question 6!
6.
A basketball player makes one out of his first two free throws. From that
point on, the probability that he makes the next shot is equal to the
proportion of shots made up to that point. If he takes two more shots, what
is the probability he ends up making a total of two free throws?
A.
1/4
B.
1/3
C.
1/2
D.
2/3
E.
3/4
Question 6 Answer
 The
Answer is B: 1/3
 P(2
shots made) = (1/2) x (1/3) + (1/2) x (1/3)