Transcript Document

Chapter 7
Probability
7.1
The Nature of Probability
Definitions



An experiment is a controlled operation that
yields a set of results.
The possible results of an experiment are called
its outcomes.
An event is a subcollection of the outcomes of
an experiment.
Definitions continued


Empirical probability is the relative frequency
of occurrence of an event and is determined by
actual observations of an experiment.
Theoretical probability is determined through
a study of the possible outcomes that can occur
for the given experiment.
Empirical Probability
number of times
event E has occurred
P(E) 
total number of times the
experiment has been performed


Example: In 100 tosses of a fair die, 19
landed showing a 3. Find the empirical
probability of the die landing showing a 3.
Let E be the event of the die landing
showing a 3.
19
P(E) 
 0.19
100
The Law of Large Numbers

The law of large numbers states that
probability statements apply in practice to a
large number of trials, not to a single trial. It is
the relative frequency over the long run that is
accurately predictable, not individual events or
precise totals.
7.2
Theoretical Probability
Equally likely outcomes


If each outcome of an experiment has the same
chance of occurring as any other outcome, they
are said to be equally likely outcomes.
For equally likely outcomes, the probability of
Event E may be calculated with the following
formula.
number of outcomes favorable to E
P(E) 
total number of possible outcomes
Example






A die is rolled. Find the probability of rolling
a) a 2.
b) an odd number.
c) a number less than 4.
d) an 8.
e) a number less than 9.
Solutions: There are six equally likely
outcomes: 1, 2, 3, 4, 5, and 6.

a)
number of outcomes that will result in a 2 1
P(2) 

total number of possible outcomes
6


b) There are three ways an odd number can
occur 1, 3 or 5.
3 1
P(odd)  
6 2
c) Three numbers are less than 4.
3 1
P(number less than 4)  
6 2
Solutions: There are six equally likely
outcomes: 1, 2, 3, 4, 5, and 6. continued

d) There are no outcomes that will result in
an 8.
0
P(number greater than 8)   0
6

e) All outcomes are less than 9. The event
must occur and the probability is 1.
Important Facts




The probability of an event that cannot occur is 0.
The probability of an event that must occur is 1.
Every probability is a number between 0 and 1
inclusive; that is, 0 ≤ P(E) ≤ 1.
The sum of the probabilities of all possible
outcomes of an experiment is 1.
Example

A standard deck of cards is well shuffled. Find
the probability that the card is selected.
a) a 10.
b) not a 10.
c) a heart.
d) an ace, 2, or 3.
e) diamond and spade.
f) a card greater than 4 and less than 7.
Example continued
a) a 10
There are four 10’s in
a deck of 52 cards.
4
1
P(10) 

52 13
b) not a 10
P(not a 10)  1 P(10)
1
 1
13
12

13
Example continued
c) a heart
There are 13 hearts
in the deck.
13 1
P(heart) 

52 4
d) an ace, 2 or 3
There are 4 aces, 4
twos and 4 threes, or
a total of 12 cards.
12 3
P(A, 1 or 2) 

52 13
Example continued
d) diamond and spade
The word and means
both events must
occur. This is not
possible.
P(diamond and spade)
0

0
52
e) a card greater than 4
and less than 7
The cards greater
than 4 and less than
7 are 5’s, and 6’s.

P 5 or 6 

P > 4 and < 7 
8
2


52 13
7.3
Odds
Odds in Favor

Odds in favor of event 

P event occurs

P success 

P failure 

P event fails to occur

Odds Against

Odds against event 

P event fails to occur

P event occurs



P success 
P failure


Example: Odds Against


Example: Find the odds against rolling a 5 on
one roll of a die.

1
P 5 
6


5
P fails to roll a 5 
6
5
5 6 5
odds against
6
   
rolling a 5
1 6 1 1
6
The odds against rolling a 5 are 5:1.
Probability from Odds



Example: The odds of spinning a blue on a
certain spinner are 4:3. Find the probability that
a) a blue is spun.
b) a blue is not spun.
Solution

Since the odds are 4:3 the denominators must
be 4 + 3 = 7.

The probabilities ratios are:


4
P blue 
7


3
P not blue 
7
7.4
Expected Value (Expectation)
Expected Value


E  P1  A1  P2  A2  P3  A3  ...  Pn  An
The symbol P1 represents the probability that
the first event will occur, and A1 represents the
net amount won or lost if the first event occurs.
Example

When Calvin Winters attends a tree farm event,
he is given the opportunity to purchase a ticket
for the $75 door prize. The cost of the ticket is
$3, and 150 tickets will be sold. Determine
Calvin’s expectation if he purchases one ticket.
Solution
 
 
1
149
E
75 
3
150
150
75 447


150 150
372

150
 2.48

Calvin’s expectation is - $2.48 when he
purchases one ticket.
7.5
Tree Diagrams
Counting Principle

If a first experiment can be performed in M
distinct ways and a second experiment can be
performed in N distinct ways, then the two
experiments in that specific order can be
performed in M • N distinct ways.
Definitions

Sample space: A list of all possible outcomes
of an experiment.

Sample point: Each individual outcome in the
sample space.

Tree diagrams are helpful in determining
sample spaces.
Example

a)
b)
c)
d)
Two balls are to be selected without
replacement from a bag that contains one
purple, one blue, and one green ball.
Use the counting principle to determine the
number of points in the sample space.
Construct a tree diagram and list the sample
space.
Find the probability that one blue ball is
selected.
Find the probability that a purple ball followed
by a green ball is selected.
Solutions
a) 3 • 2 = 6 ways
b)
B PB
P
G PG
BP
B
P
G
G
P
BG
GP
B
GB


c)
4 2
P blue  
6 3
d)
P Purple,Green

1
P P,G
6

P(event happening at least once)
 event happening
 event does 
P
 1 P 

 at least once

 not happen
7.6
Or and And Problems
Or Problems


P(A or B) = P(A) + P(B) - P(A and B)
Example: Each of the numbers 1, 2, 3, 4, 5, 6,
7, 8, 9, and 10 is written on a separate piece of
paper. The 10 pieces of paper are then placed
in a bowl and one is randomly selected. Find
the probability that the piece of paper selected
contains an even number or a number greater
than 5.
Solution

P(A or B) = P(A) + P(B) - P(A and B)
 even or

P


 greater than 5
 even and

P even  P greater than 5  P 
 greater than 5

 

5
5
3
7




10 10 10 10

Thus, the probability of selecting an even
number or a number greater than 5 is 7/10.
Example

Each of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, and
10 is written on a separate piece of paper. The
10 pieces of paper are then placed in a bowl
and one is randomly selected. Find the
probability that the piece of paper selected
contains a number less than 3 or a number
greater than 7.
Solution


2
P less than 3 
10


3
P greater than 7 
10
There are no numbers that are both less than 3
and greater than 7. Therefore,
 less than 3 or  2
3
5 1
P


0


10 2
 greater than 7 10 10
Mutually Exclusive

Two events A and B are mutually exclusive if
it is impossible for both events to occur
simultaneously.
Example

One card is selected from a standard deck of
playing cards. Determine the probability of the
following events.
a) selecting a 3 or a jack
b) selecting a jack or a heart
Solutions
a) 3 or a jack
 

4
4
P 3  P jack 

52 52
8
2


52 13
b) jack or a heart
 jack and
4 13 1
P jack  P heart  P 




 heart
 52 52 52

 

16 4


52 13
And Problems


P(A and B) = P(A) • P(B)
Example: Two cards are to be selected with
replacement from a deck of cards. Find the
probability that two red cards will be selected.
     
P A  P B  P red  P red
26 26


52 52
1 1 1
  
2 2 4
Example

Two cards are to be selected without
replacement from a deck of cards. Find the
probability that two red cards will be selected.
P  A   P  B   P  red   P  red 
26 25


52 51
1 25 25
 

2 51 102
Independent Events

Event A and Event B are independent events
if the occurrence of either event in no way
affects the probability of the occurrence of the
other event.

Experiments done with replacement will result in
independent events, and those done without
replacement will result in dependent events.