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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
Important Terms
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
Another Experiment
Probability Experiment: An action through
which counts, measurements, or responses are
obtained
Choose a car from production line
Sample Space: The set of all possible outcomes
Event: A subset of the sample space.
Outcome: The result of a single trial
Larson/Farber Ch. 3
Types of Probability
Classical (equally probable outcomes)
Empirical
Probability blood pressure will decrease
after medication
Intuition
Probability the line will be busy
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
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
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.
E
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?
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 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
Independent Events
Two events A and B are independent if the
probability of the occurrence of event B is not
affected by the 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
Larson/Farber Ch. 3
A = being a female
B = being under 64” tall
Independent 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.
Two dice are rolled. A = first is a 4 and B = second is a 4
P(B) = 1/6 and P(B|A) = 1/6. The events are
independent.
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)
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 = 0.1515
Larson/Farber Ch. 3
Multiplication Rule
Two dice are rolled. Find the probability both are 4’s.
A = first die is a 4 and B = second die is a 4.
P(A) = 1/6
P(B|A) = 1/6
P(A and B) = 1/6 x 1/6 = 1/36 = 0.028
When two events A and B are independent, then
P (A and B) = P(A) x P(B)
Note for independent events P(B) and P(B|A) are
the same.
Larson/Farber Ch. 3
Section 3.3
The Addition Rule
Larson/Farber Ch. 3
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
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 West Wing on TV
A and B
Non-mutually exclusive
P(A and B) ≠ 0
Larson/Farber Ch. 3
A
B
The Addition Rule
The probability that one or the other of two events will
occur is:
P(A) + P(B) – P(A and B)
A card is drawn from a 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) = 4/52 and P(B) = 26/52
but 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 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 = 0.054
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 = 150/1000
= 0.15
2. P(Miami and Seattle) = 0
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
3 P(Miami or Yes)
250/1000 + 400/1000 – 150/1000
= 500/1000 = 0.5
4. P(Miami or
Seattle)
250/1000 + 450/1000 – 0/1000
= 700/1000 = 0.7
Larson/Farber Ch. 3
Summary
For complementary events P(E') = 1 - P(E)
Subtract the probability of the event from one.
The probability both of two events occur
P(A and B) = P(A) • P(B|A)
Multiply the probability of the first event by the conditional probability
the second event occurs, given the first occurred.
Probability at least one of two events occur
P(A or B) = P(A) + P(B) - P(A and B)
Add the simple probabilities, but to prevent double counting, don’t
forget to subtract the probability of both occurring.
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, the 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.
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
Factorials
Suppose you want to arrange n objects in
order.
There are n choices for 1st place.
Leaving n – 1 choices for second, then n – 2 choices for
third place and so on until there is one choice of 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
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 (where r £ n) is:
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
Combinations
A combination is a selection of r objects from a group o
f n objects.
The number of combinations of n objects taken r at a time is
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