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Probability
Psychology
Weather forecast
Business
Elementary Statistics
Larson
Farber
Games
Medicine
Sports
Larson/Farber Ch. 3
Section 3.1
Basic Concepts of
Probability
Larson/Farber Ch. 3
Example
Probability experiment:
Roll a die
An action through which counts, measurements or
responses are obtained
Sample space:
{1 2 3 4 5 6}
The set of all possible outcomes
Event: { Die is even } = { 2 4 6 }
A subset of the sample space.
Outcome:
{4}
The result of a single trial
Larson/Farber Ch. 3
Simple Event
Simple Event – an event that consists of
a single outcome
Example:
“tossing heads” and “rolling a 3” is a simple
event because there is 1 possible outcome.
“tossing heads and “rolling an even number”
is not a simple event because there are 3
possible outcomes
Larson/Farber Ch. 3
Types of Probability
Classical (theoretical)
Empirical (actual, statistical, experimental)
Subjective (intuition)
Based on educated guesses, intuition and estimates
Larson/Farber Ch. 3
Law of Large Numbers
As you increase the number of times
a probability experiment is repeated,
the empirical probability of the event
approaches the theoretical probability
of the event.
Larson/Farber Ch. 3
Tree Diagrams
Two dice are rolled.
Describe the
sample space.
1st roll
1
2
1 2 3 4 5 6
Start
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
2nd roll
36 outcomes
Larson/Farber Ch. 3
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
5,1
5,2
5,3
5,4
5,5
5,6
Find the probability the sum is 4.
Find the probability the sum is 11.
Find the probability the sum is 4 or 11.
Larson/Farber Ch. 3
6,1
6,2
6,3
6,4
6,5
6,6
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
5,1
5,2
5,3
5,4
5,5
5,6
6,1
6,2
6,3
6,4
6,5
6,6
Find the probability the sum is 4.
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.
Larson/Farber Ch. 3
5/36 = 0.139
Range of Probabilities
The probability of an event E
is between 0 and 1, inclusive
Probability cannot be negative
Probability cannot be greater than 1
If probability is 1, the event is certain
If probability is 0, the event is impossible
Larson/Farber Ch. 3
Complementary Events
The complement of event E is event E´. E´ consists
of all the events in the sample space that are not in
event E.
P(E´) = 1 - P(E)
The day’s production consists of 12 cars, 5 of
which are defective. If one car is selected at
random, find the probability it is not defective.
Solution:
P(defective) = 5/12
P(not defective) = 1 - 5/12 = 7/12 = 0.583
Larson/Farber Ch. 3
Section 3.2
Conditional Probability
and the
Multiplication Rule
Larson/Farber Ch. 3
Conditional Probability
The probability an event B will occur, given (on the
condition) that another event A has occurred.
We write this as P(B|A) and say “probability of B, given A.”
Two cars are selected from a production line of
12 cars where 5 are defective. What is the
probability the 2nd car is defective, given the first
car was defective?
Larson/Farber Ch. 3
Conditional Probability
Two cars are selected from a production line of
12 cars where 5 are defective. What is the
probability the 2nd car is defective, given the first
car was defective?
Given a defective car has been selected, the
conditional sample space has 4 defective out of 11.
P(B|A) = 4/11
Larson/Farber Ch. 3
Independent / Dependent Events
Two events A and B are independent if the
probability of the occurrence of event B is not
affected by the 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
Larson/Farber Ch. 3
A = being a female
B = being under 64” tall
Independent Events
Two dice are rolled. Find the probability
the second 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
Larson/Farber Ch. 3
Dependent Events
If events A and B are independent, then P(B|A) = P(B)
Conditional Probability
Probability
12 cars are on a production line where 5 are defective and 2
cars are selected at random.
A = first car is defective
B = second car is defective.
The probability of getting a defective car for the second car
depends on whether the first was defective. The events are
dependent.
Larson/Farber Ch. 3
Contingency Table
The results of responses when a sample of adults
in 3 cities was asked if they liked a new juice is:
Omaha
Yes
100
No
125
Undecided
75
Total
300
Seattle
150
130
170
450
Miami
150
95
5
250
Total
400
350
250
1000
One of the responses is selected at random. Find:
1. P(Yes)
2. P(Seattle)
3. P(Miami)
4. P(No, given Miami)
Larson/Farber Ch. 3
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
Answers: 1) 0.4
Larson/Farber Ch. 3
Total
400
350
250
1000
2) 0.45
3) 0.25
4) 0.38
Multiplication Rule
To find the probability that two events, A and B will occur in
sequence, multiply the probability A occurs by the
conditional probability B occurs, given A has occurred.
P(A and B) = P(A) x P(B|A)
If events A and B are independent then the rule can be
simplified to
P(A and B) = P(A) x P(B)
Larson/Farber Ch. 3
Multiplication Rule
Two 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
B = second car is defective.
P(A) = 5/12
P(B|A) = 4/11
P(A and B) = 5/12 x 4/11 = 5/33 = 0.152
Larson/Farber Ch. 3
Multiplication Rule
Two dice are rolled. Find the probability both are 4’s.
Larson/Farber Ch. 3
Multiplication Rule
Two dice are rolled. Find the probability both are 4’s.
When two events A and B are independent, then
P (A and B) = P(A) x P(B)
P(A and B) = 1/6 x 1/6 = 1/36 = 0.028
Larson/Farber Ch. 3
Section 3.3
The Addition Rule
Larson/Farber Ch. 3
Mutually Exclusive Events
Two events, A and B, are mutually exclusive if
they cannot occur in the same trial.
A = A person is under 21 years old
B = A person is running for the U.S. Senate
A = A person was born in Philadelphia
B = A person was born in Houston
A
B
Mutually exclusive
P(A and B) = 0
When event A occurs it excludes event B in the same trial.
Larson/Farber Ch. 3
Non-Mutually Exclusive Events
If two events can occur in the same trial, they are
non-mutually exclusive.
A = A person is under 25 years old
B = A person is a lawyer
A = A person was born in Philadelphia
B = A person watches Revenge on TV
A and B
Non-mutually exclusive
P(A and B) ≠ 0
Larson/Farber Ch. 3
A
B
Compare “A and B” to “A or B”
The compound event “A and B” means that A
and B both occur in the same trial. Use the
multiplication rule to find P(A and B).
The compound event “A or B” means either A
can occur without B, B can occur without A or
both A and B can occur. Use the addition rule
to find P(A or B).
B
A
A and B
Larson/Farber Ch. 3
B
A
A or B
The Addition Rule
The probability that one or the other of two events will
occur is:
P(A or B) = P(A) + P(B) – P(A and B)
Larson/Farber Ch. 3
The Addition Rule
A card is drawn from a standard deck. Find
the probability it is a king or it is red.
A = the card is a king
B = the card is red.
P(A or B) = P(A) + P(B) – P(A and B)
P(A) = 4/52, P(B) = 26/52, P(A and B) = 2/52
P(A or B) = 4/52 + 26/52 – 2/52 = 28/52 = 0.538
Larson/Farber Ch. 3
The Addition Rule
A card is drawn from a standard deck. Find
the probability the card is a king or a 10.
Larson/Farber Ch. 3
The Addition Rule
A card is drawn from a standard deck. Find
the probability the card is a king or a 10.
A = the card is a king
B = the card is a 10.
P(A) = 4/52 and P(B) = 4/52 and P(A and B) = 0/52
P(A or B) = 4/52 + 4/52 – 0/52 = 8/52 = 0.154
When events are mutually exclusive,
P(A or B) = P(A) + P(B)
Larson/Farber Ch. 3
Contingency Table
The results of responses when a sample of adults in
3 cities was asked if they liked a new juice is:
Omaha
Yes
100
No
125
Undecided 75
Total
300
Seattle
150
130
170
450
Miami
150
95
5
250
Total
400
350
250
1000
One of the responses is selected at random. Find:
1. P(Miami and Yes)
3. P(Miami or Yes)
2. P(Miami and Seattle) 4. P(Miami or Seattle)
Larson/Farber Ch. 3
Contingency Table
Yes
No
Undecided
Total
Omaha
100
125
75
300
Seattle
150
130
170
450
Miami
150
95
5
250
Total
400
350
250
1000
One of the responses is selected at random. Find:
1. P(Miami and Yes) = 250/1000 • 150/250 = 0.15
2. P(Miami and Seattle) = 0
3. P(Miami or Yes) = 250/1000 + 400/1000 – 150/1000 = 0.5
4. P(Miami or Seattle) = 250/1000 + 450/1000 – 0/1000 = 0.7
Larson/Farber Ch. 3
Summary
See page 144 in the text book for a
summary of probability
Larson/Farber Ch. 3
Section 3.4
Counting Principles
Larson/Farber Ch. 3
Fundamental Counting Principle
If one event can occur m ways and a second
event can occur n ways, then number of ways
the two events can occur in sequence is m • n.
This rule can be extended for any number of
events occurring in a sequence.
Larson/Farber Ch. 3
Fundamental Counting Principle
If a meal consists of 2 choices of soup, 3 main dishes and
2 desserts, how many different meals can be selected?
Dessert
Soup
Main
Start
2
Larson/Farber Ch. 3
•
3
•
2
= 12 meals
Fundamental Counting Principle
How many license plates can you make if a license
plate consists of
a. 6 letters which can be repeated
b. 6 letters which cannot be repeated
Larson/Farber Ch. 3
Factorials
Suppose you want to arrange n objects in order.
There are n choices for 1st place, n – 1 choices
for second, then n – 2 choices for third place and
so on until there is one choice for last place.
Using the Fundamental Counting Principle, the
number of ways of arranging n objects is:
n(n – 1)(n – 2)…1
This is called n factorial and written as n!
Larson/Farber Ch. 3
Factorials
The starting lineup for a baseball team
consists of nine players. How many
different batting orders are possible using
the starting lineup?
Larson/Farber Ch. 3
Permutations
A permutation is an ordered arrangement.
The number of permutations for n objects is n!
n! = n (n – 1) (n – 2)…..3 • 2 • 1
The number of permutations of n objects
taken r at a time is:
Larson/Farber Ch. 3
Permutations
You are required to read 5 books from a list of 8. In how many
different orders can you do so?
There are 6720 permutations of 8 books reading 5.
Larson/Farber Ch. 3
Distinguishable Permutations
n!
n1 !n2 !n3 !..........nk !
where n1+n2+n3+…….+nk = n
A contractor wants to plant 6 oak trees, 9 maple trees
and 5 poplar trees along the street. In how many
distinguishable ways can they be planted?
Larson/Farber Ch. 3
Combinations
A combination is a selection of r objects from a group
of n objects.
The number of combinations of n objects taken r at a time is:
Larson/Farber Ch. 3
Combinations
You are required to read 5 books from a list of 8. In how
many different ways can you choose the books if order does
not matter.
There are 56 combinations of 8 objects taking 5.
Larson/Farber Ch. 3
1
2
3
4
Combinations of 4 objects choosing 2
1
2
3
1
1
4
2
3
3
4
2
4
Each of the 6 groups represents a combination.
Larson/Farber Ch. 3
1
2
4
3
Permutations of 4 objects choosing 2
1
1
1
2
3
2
3
4
4
1
1
1
2
3
3
4
3
4
2
3
2
4
4
2
Each of the 12 groups represents a permutation.
Larson/Farber Ch. 3
Applications
A word consists of one L, two Es, two Ts, and one R.
If the letters are randomly arranged in order, what is the
probability that the arrangement spells the word letter?
Larson/Farber Ch. 3
Applications
Find the probability of being dealt five diamonds from a
standard deck of playing cards.
Larson/Farber Ch. 3