Probability, Part 1
We’ve looked at several counting
techniques to determine the number
of possible arrangements or
groupings in a given scenario. These
techniques include the multiplication
principle, the addition principle,
permutations, and combinations.
Probability, Part 1
Together these techniques provide us
with a powerful set of tools for
calculating probabilities.
Probability, Part 1
Probability tells us the likelihood of a
given event occurring.
We normally represent it as a decimal
or fraction between 0 and 1, or as a
percentage between 0% and 100%.
Probability, Part 1
For example, a weather reporter
might say there is a 70% chance, or
probability, of rain tomorrow.
We could also say
P(rain) = 0.7
Probability, Part 1
A probability of 1, or 100%, indicates
certainty that a given even will occur.
A probability of 0, or 0%, indicates
certainty that a given even will not
occur.
Probability, Part 1
Example 1: Say you roll a die. What
is the probability that you will get a
number between 1 and 6? What is the
probability you will get a 7?
P(1-6) = 1, because we are
guaranteed to get one of the numbers
from 1 to 6.
Probability, Part 1
Example 1: Say you roll a die. What
is the probability that you will get a
number between 1 and 6? What is the
probability you will get a 7?
P(7) = 0, because it is impossible to
get a 7 with only one die.
Probability, Part 1
This example was rather simple, but the
problems can become much more complex. We
need to have a more systematic approach to
handle those complex problems.
In general, we will want to determine the
number of possible outcomes that fit a set
of conditions, and compare that to the total
possible number of outcomes.
Probability, Part 1
In general, the probability of a given event
occurring is equal to the ratio:
__Number of Outcomes That Work__
Total Possible Number of Outcomes
Probability, Part 1
Example 2: If you roll a die, what is the
probability of getting an even number?
____Number of Even Outcomes____
Total Possible Number of Outcomes
Probability, Part 1
Example 2: If you roll a die, what is the
probability of getting an even number?
Possible Outcomes: 1, 2, 3, 4, 5, 6
So there are 6 total possible outcomes.
Probability, Part 1
Example 2: If you roll a die, what is the
probability of getting an even number?
Even Outcomes: 2, 4, 6
There are 3 even outcomes.
Probability, Part 1
Example 2: If you roll a die, what is the
probability of getting an even number?
3
1
P(even) =
=
6
2
Probability, Part 1
Again, this example was rather simple.
Let's look at a couple other examples that
require more thought.
Probability, Part 1
Example 3: You are taking a 5-question
matching quiz; you are asked to match 5
vocabulary terms with 5 definitions. Alas,
you are ill-prepared for the quiz, and end
up totally guessing on the whole thing.
What is the probability that by guessing
you will score 100%?
Probability, Part 1
Example 3: We can view this as a
permutation problem to determine the
total possible number of ways to answer
the quiz. In a sense, you are being asked
to arrange 5 definitions in order. The
number of ways to do this is
P(5, 5) = 5 nPr 5 = 120.
Probability, Part 1
Example 3: The outcomes that “work” in
this case are the outcomes that will give
you a score of 100%. There is only 1 such
outcome.
Hence, the probability of scoring 100% is
1 =
120
0.0083, or about 0.83%.
Probability, Part 1
Example 4: You are taking a 10 question
True/False quiz. You are so unsure of the
questions that you decided to flip a coin to
help you answer each one. (If you get
heads, you answer true; tails you answer
false.) In other words, you're answering
completely at random. What is the
probability you will score exactly 70%?
Probability, Part 1
Example 4: 10 question True/False Quiz
First, let's figure out the total number of ways to
answer the quiz. The first question has two
options: true or false. So does the second
question, and the third, ...
So the total possible number of outcomes is
2*2*2*2*2*2*2*2*2*2 = 210 = 1,024.
Probability, Part 1
Example 4: 10 question True/False Quiz
Now let's figure out the number of outcomes that
“work,” or in this case that give you a score of
70%. Assuming there is no partial credit, this
means getting 7 out of the 10 questions correct.
It doesn't matter which 7 questions you get
correct, though—it could be 1-7, 4-10, 1-5 and
8-9, etc.
Probability, Part 1
Example 4: 10 question True/False Quiz
In other words, we need the number of
combinations of 7 correct answers out of 10
problems. (Why combinations rather than
permutations?) The number of ways to score
70% is
C(10, 7) = 10 nCr 7 = 120.
Probability, Part 1
Example 4: 10 question True/False Quiz
The probability of getting 70% is then
__120__
1,024
= 0.117, or about 11.7%.
Probability, Part 1
In the previous example we used the
multiplication principle together with
combinations. In some instances we will
use the addition principle in concert with
combinations or permutations.
(Recall that a key word for the addition
principle is frequently “or.”)
Probability, Part 1
Example 5: You are again completely
guessing on a 10 question True/False quiz.
What is the probability that you will score
at least 70%?
Probability, Part 1
Example 5: 10 question True/False Quiz
Again assume that no partial credit is
given. In this case, the outcomes that
“work” are scores of 70% or 80% or 90%
or 100%. We need to determine these
possibilities and then add them together.
Probability, Part 1
Example 5: 10 question True/False Quiz
We already calculated the number of ways
to get 7 out of 10 questions correct:
C(10, 7) = 10 nCr 7 = 120.
Probability, Part 1
Example 5: 10 question True/False Quiz
Consider also the number of ways to get 8,
9, or 10 correct:
C(10, 8) = 10 nCr 8 = 45.
C(10, 9) = 10 nCr 9 = 10.
C(10, 10) = 10 nCr 10 = 1.
Probability, Part 1
Example 5: 10 question True/False Quiz
Now add together the number of ways to
get 7, 8, 9, or 10 questions correct.
120 + 45 + 10 + 1 = 176
This gives us a total of 176 outcomes that work.
Probability, Part 1
Example 5: 10 question True/False Quiz
The total possible number of ways to
answer the quiz remains 1,024, as we
already calculated in Example 4.
Probability, Part 1
Example 5: 10 question True/False Quiz
This gives us a probability of scoring at
least 70% on the quiz as:
176__
= 0.172, or about 17.2%.
1,024
Probability, Part 1
The Addition Principle for Mutually Exclusive Events
When considering 2 events, A and B, we can find
the probability of one or the other occurring by
simply adding together their individual
probabilities, with one adjustment.
Probability, Part 1
The Addition Principle for Mutually Exclusive Events
When considering 2 events, A and B, we can find
the probability of one or the other occurring by
simply adding together their individual
probabilities, with one adjustment.
We need to subtract the overlap between the 2 events.
P(A or B) = P(A) + P(B) – P(A and B)
Probability, Part 1
The Addition Principle for Mutually Exclusive Events
In a case where A and B are mutually exclusive,
both cannot occur at the same time. As a result,
P(A and B) = 0, and our equation simplifies to
P(A or B) = P(A) + P(B)
Probability, Part 1
The Addition Principle for Mutually Exclusive Events
Confused yet?
Let's look at some examples to help illustrate.
Probability, Part 1
The Addition Principle for Mutually Exclusive Events
Example 6: A car is being given away at a high
school graduation. What is the probability that a
football player or band member will win the car,
if among the 400 graduates there are 40 football
players and 50 band members, including 5 that
participated in both football and band?
Probability, Part 1
The Addition Principle for Mutually Exclusive Events
Example 6: Car Giveaway
Out of 40 football players, 5 were also in band,
so there were 35 who played football but were
not in band.
Out of 50 band members, 5 also played football,
so there were 45 who were in band but did not
play football.
Probability, Part 1
The Addition Principle for Mutually Exclusive Events
Example 6: Car Giveaway
Band
Football
35
5
45
Probability, Part 1
The Addition Principle for Mutually Exclusive Events
Example 6: Car Giveaway
Band
Football
35
5
45
P(Football or Band) = P(Football) + P(Band) – P(Football and Band)
= 40_ + 50_ - 5 = 85
400
400
400
400
= 17
80
Probability, Part 1
The Addition Principle for Mutually Exclusive Events
Example 7: At the same high school graduation,
what is the probability that a football player or
cross country runner will win the car, if among
the 400 graduates there are 40 football players
and 50 cross country runners, and no one
participated in both?
Probability, Part 1
The Addition Principle for Mutually Exclusive Events
Example 7: Car Giveaway
Football
40
P(Football or CC) = P(Football) + P(CC)
= 40_ + 50_ = 90 = 9
400
400
400
40
CC
50
Probability, Part 1
Conditional Probability:Using the Multiplication Principle
We can use the multiplication principle to help
determine the probability of two events both
occurring (and rather than or):
P(A and B) = P(A) * P(B from A)
Once again, this may be confusing without an example.
Probability, Part 1
Conditional Probability:Using the Multiplication Principle
Example 8: A certain high school has the following
student population:
Female
Male
Total
Fr.
123
105
228
Soph.
168
132
300
Jr.
145
110
255
Sr.
124
93
217
Total
560
440
1000
What is the probability that a student selected at random
will be a male junior?
Probability, Part 1
Conditional Probability:Using the Multiplication Principle
Example 8: P(Junior and Male)
Consider being a junior as Event A, and being
male as Event B. We’re trying to determine
P(A and B).
We could do this by constructing a tree diagram.
Probability, Part 1
Conditional Probability:Using the Multiplication Principle
Example 8: P(Junior and Male)
.228
.3
Fr
So
.255
Jr
.217
Sr
Label the probabilities for
each grade level. These
come from the number of
students in each grade
divided by the total number
of students.
E.g., P(So) = 300/1000 = .3
Probability, Part 1
Conditional Probability:Using the Multiplication Principle
Example 8: P(Junior and Male)
.228
Fr
.461 M
.539
F
.560 M
.3
So
.440
F
.255
.431 M
Jr
.217
.569
F
Sr
.423 M
.577
F
Label the
probabilities
of being male
or female
within each
grade level.
Probability, Part 1
Conditional Probability:Using the Multiplication Principle
Example 8: P(Junior and Male)
.228
Fr
.461 M
.539
F
.560 M
.3
So
.440
F
.255
.431 M .110
Jr
Follow the path for
Juniors and then
Males, and multiply
the probabilities.
.217
.569
F
Sr
.423 M
.577
F
.255*.431
= .110
Probability, Part 1
Conditional Probability:Using the Multiplication Principle
Example 8: P(Junior and Male)
We can get the same result by using the formula:
P(A and B) = P(A) * P(B from A)
Probability, Part 1
Conditional Probability:Using the Multiplication Principle
Example 8: P(Junior and Male)
P(A and B) = P(A) * P(B from A)
P(Junior and Male) = P(Junior) * P(Male from Juniors)
P(Junior) = 255/1000 = .255
P(Male from Junior) = 110/255 = .431
P(Junior and Male) = .255 * .431 = .110
Probability, Part 1
Conditional Probability:Using the Multiplication Principle
Our formula for conditional probabilities becomes a
small bit simpler when the events are independent. In
this case:
P(A and B) = P(A) * P(B)
Probability, Part 1
Conditional Probability:Using the Multiplication Principle
Two events are independent if
P(B from A) = P(B)
or
P(A from B) = P(A)
Probability, Part 1
Conditional Probability:Using the Multiplication Principle
Consider again our example of the school where Event
A is being a Junior, and Event B is being a male.
Does P(B from A) = P(B)?
Probability, Part 1
Conditional Probability:Using the Multiplication Principle
Consider again our example of the school where Event
A is being a Junior, and Event B is being a male.
P(B) = P(Male) = 440/1000 = .440
P(B from A) = P(Male from Juniors) = 110/255 = .431
So P(B from A)  P(B)
Probability, Part 1
Conditional Probability:Using the Multiplication Principle
Consider again our example of the school where Event
A is being a Junior, and Event B is being a male.
P(A) = P(Junior) = 255/1000 = .255
P(A from B) = P(Junior from Males) = 110/440 = .250
So P(A from B)  P(A)
Being a junior and being male are therefore not
independent.
Probability, Part 1
Conditional Probability:Using the Multiplication Principle
Example 9: Are being a sophomore and being female
independent in the school of Example 8?
Female
Male
Total
Fr.
123
105
228
Soph.
168
132
300
Jr.
145
110
255
Sr.
124
93
217
Total
560
440
1000
Probability, Part 1
Conditional Probability:Using the Multiplication Principle
Example 9: Are being a sophomore and being female
independent in the school of Example 8?
P(B) = P(Female) = 560/1000 = .560
P(B from A) = P(Female from Sophomores) = 168/300
= .560
Because P(B from A) = P(B), being a sophomore and
being female are independent in this school.
Probability, Part 1
Conditional Probability:Using the Multiplication Principle
Example 9: Are being a sophomore and being female
independent in the school of Example 8?
In other words, the probability of picking a female from
among the sophomores is the same as the probability of
picking a female from the school population at large.
In contrast, the probability of picking a male from
among the juniors is slightly less than the probability of
picking a male from the school population at large.
Probability, Part 1
Conditional Probability:Using the Multiplication Principle
Example 10: What is the probability that a student
chosen at random from the entire school population will
be a female sophomore?
Probability, Part 1
Conditional Probability:Using the Multiplication Principle
Example 10: What is the probability that a student chosen at
random from the entire school population will be a female
sophomore?
We already established that being female and being a
sophomore are independent events in this case, so we
can use our simplified formula:
P(Female and Sophomore) = P(F) * P(Soph)
= .560 * .300 = .168
This makes sense as there are 168 female sophomores out of
1,000 total students in the school.