Section 3.1
Basic Concepts of
Probability
Definitions
Probability experiment - An action through which
counts, measurements, or responses are obtained.
Example: roll a die
Sample space – the set of all possible outcomes
Example: {1 2 3 4 5 6}
Event – a subset of the sample space
Example: {Die is even} = {2 4 6}
Outcome – the result of a single trial
Example: {4}
Another Example
Probability Experiment:
choose a car from an assembly line
Sample Space: all cars on the assembly line
Event: every 4th car on the assembly line
Outcome: bright, shiny blue truck
Fundamental Counting Principle
If too many ways to effectively count all possible
outcomes, then how to determine number of ways
events can occur in sequence?
If one event can occur in m ways and a 2nd event can
occur in n ways,
Then the number of ways that the two events can occur
in sequence is m x n (not restricted to 2 events)
Types of Probability
Classical or Theoretical – assumes equally probable
outcomes
Empirical or Statistical - based on relative frequency
of event E
f

n
Example: probability that blood pressure will decrease after
medication
Intuition or Guess – based on past experience
Example: probability that the phone line will be busy
Tree Diagrams
Two dice are rolled.
Describe the sample space.
Start
1st roll
1
1 2 3 4 5 6
2nd roll
2
3
4
5
6
1 2 3 4 5 6 12 3 4 5 61 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6
Sample space: 36 possible outcomes
Use Fundamental Counting Principle?
mxn=6x6
Sample Space and Probabilities
Two dice are rolled and the sum is noted.
1,1
1,2
1,3
1,4
1,5
1,6
2,1
2,2
2,3
2,4
2,5
2,6
3,1
3,2
3,3
3,4
3,5
3,6
4,1
4,2
4,3
4,4
4,5
4,6
Find the probability the sum is 4.
5,1
5,2
5,3
5,4
5,5
5,6
6,1
6,2
6,3
6,4
6,5
6,6
3/36 = 1/12 = 0.083
Find the probability the sum is 11. 2/36 = 1/18 = 0.056
Find the probability the sum is 4 or 11.
5/36 = 0.139
Class Experiment
1. Each student toss their penny 10 times
2. Record the outcome of each toss
3. Calculate the proportion of heads for each of the
10 tosses
4. Predict the trend, plot the result
Probability of Tossing a Head
1.0
Proportion heads
0.8
0.6
0.4
0.2
0.0
0
1
2
3
4
5
6
7
Number of tosses
8
9
10
11
Large Sample Sets
Law of Large Numbers – as sample size increases, the
empirical probability of an event approaches the theoretical
(actual) probability of the event.
Example: getting a head
when tossing a coin
Range of Probabilities Rule – probability of an event E
is between 0 and 1:
0  PE   1
Complementary Events
Complement of event E - event E´. E´ is all events in
the sample space that are not in event E.
P(E´) = 1 - P(E)
Example: Factory day’s production = 12 cars, 5 are
defective. If one car is selected at random, find the probability
it is not defective.
P(defective) = 5/12
P(not defective) = 1 - 5/12 = 7/12 = 0.583
Section 3.2
Conditional Probability and the
Multiplication Rule
Conditional Probability
Probability that event B will occur, given that event
A has occurred.
P(B|A) = “probability of B, given A.”
Example: 2 cars are selected from 12 cars where 5 are
defective. What is the probability the 2nd car is defective,
given the first car was defective?
Probability that first car is defective (conditional?):
P(defective) = 5/12 = 0.417
Given a defective car has been selected, the conditional sample
space has 4 defective out of 11. P(B|A) = 4/11 = 0.363
Independent Events
A and B are independent if the probability of event B
is not affected by occurrence (or non-occurrence) of
event A.
A = Being female
B = Having type O blood
A = 1st child is a boy
B = 2nd child is a boy
Two events that are not independent are dependent.
A = taking an aspirin each day
B = having a heart attack
A = being a female
B = being under 64” tall
Independent vs Dependent
If events A and B are independent, then P(B|A) = P(B)
Conditional Probability
Probability
Example: 12 cars are on a production line, 5 are defective
and 2 cars are selected at random.
A = first car is defective
B = second car is defective
Probability of getting a defective 2nd car depends on whether
the 1st was defective. The events are dependent.
Two dice are rolled. Are the events dependent or independent?
Independent Example
Independence – occurrence of first event does not affect
probability of occurrence of second event.
PB A  PB 
PA B   P A
Example: 2 dice are rolled. Find the probability the 2nd die
is a 4 given the first was a 4.
Original sample space: {1, 2, 3, 4, 5, 6}
Given the first die was a 4, the conditional sample space is:
{1, 2, 3, 4, 5, 6}
The conditional probability, P(B|A) = 1/6
Contingency Table
Contingency or Cross Tabulation Table – records
relationships between 2 or more categorical variables.
Example: Responses of adults in 3 cities when asked if
they liked a new juice was:
Omaha
Yes
100
No
125
Undecided
75
Total
300
Seattle
150
130
170
450
Miami
150
95
5
250
One response is selected at random. Find:
1. P(Yes)
3. P(Miami)
2. P(Seattle)
4. P(No, given Miami)
Total
400
350
250
1000
Solutions
Yes
No
Undecided
Total
Omaha
100
125
75
300
1. P(Yes)
Seattle
150
130
170
450
Miami
150
95
5
250
= 400 / 1000 = 0.4
2. P(Seattle)
= 450 / 1000 = 0.45
3. P(Miami)
= 250 / 1000 = 0.25
4. P(No, given Miami) = 95 / 250 = 0.38
Total
400
350
250
1000
Multiplication Rule
Probability that 2 events, A and B, will occur in sequence,
multiply the probability of A occurring by the conditional
probability of B occurring, given that A has occurred.
P(A and B) = P(A) x P(B|A)
Example: 2 cars are selected from a production line of
12 where 5 are defective. Find the probability both cars are
defective.
A = first car is defective
P(A) = 5/12
B = second car is defective.
P(B|A) = 4/11
P(A and B) = 5/12 x 4/11 = 5/33 = 0.1515
Multiplication Rule
For independent events P(B) and P(B|A) are the same
P (A and B) = P(A) x P(B)
Example: 2 dice are rolled. Find the probability both are 4’s.
A = first die is a 4
P(A) = 1/6
B = second die is a 4
P(B|A) = 1/6
P(A and B) = 1/6 x 1/6 = 1/36 = 0.028
Probability Example
Consider a population of students. Let M be the event “a person is male” and Z be
the event that a person is interested in statistics. Suppose: P(M) = 0.4, P(Z) =
0.2, and P(M|Z) = 0.5. Find:
1) The probability of a student being male and being interested in statistics.
2) The probability of a student being interested in statistics given that they are
male.
3) The probability of a student not being male and being interested in statistics.
4) The probability of a student being interested in statistics given that they are
not male.
Steps to solve problem:
1. Figure out dimensions and draw a contingency table
2. Fill in known values
3. Arithmetic for unknown values to complete table
4. Check row and column totals
5. Answer question(s)
Contingency Table
2x2
Male
Interested in Stat
Not Interested in Stat
Total
Not Male
Total
Contingency Table
Male
Not Male
Total
Interested in Stat
0.1
0.1
0.2
Not Interested in Stat
0.3
0.5
0.8
Total
0.4
0.6
1.0
1) P(MZ) = 0.1
(male and stat)
2) P(ZM) = 0.1/0.4 = 0.25
(stat given male)
3) P(M’Z) = 0.1
(not male and stat)
4) P(ZM’) = 0.1/0.6 = 0.167
(stat given not male)