Transcript Chapter 7

Chapter 7 Probability
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7.1 Experiments, Sample Spaces, and Events
7.2 Definition of Probability
7.3 Rules of Probability
7.4 Use of Counting Techniques in Probability
7.5 Conditional Probability and Independent Events
7.6 Bayes’ Theorem
Section 7.1 Experiments, Sample Spaces, and Events
An experiment is an activity with observable results (called
outcomes).
A sample point is an outcome of an experiment. The sample space is
the set of all possible sample points.
An event is a subset of a sample space.
Ex. Rolling a die
Outcomes: landing with a 1, 2, 3, 4, 5, or 6 face up.
Sample Space: S ={1, 2, 3, 4, 5, 6}
Events: , {1}, {2}, {3}, {4}, {5}, {6}, S
Impossible
event
Certain
event
Ex. An experiment consists of spinning the hand on the disk below
two times. Find the sample space.
C
P
W
S = {(P,C), (P,W), (P,P), (C,P), (C,W), (C,C), (W,P), (W,C), (W,W)}
Events
The union of events A and B is the event A  B.
The intersection of events A and B is the event A  B.
The complement of event A is the event AC.
Ex.
Rolling a die. S = {1, 2, 3, 4, 5, 6}
Let A = {1, 2, 3} and B = {1, 3, 5}
A  B  {1, 2,3,5}
A  B  {1,3}
A  BC  {2}
Events A and B are mutually exclusive if A  B  .
Ex. When rolling a die, if event A = {2, 4, 6} (evens) and event B = {1,
3, 5} (odds), then A and B are mutually exclusive.
Ex. When drawing a single card from a standard deck of cards, if event
A = {heart, diamond} (red) and event B = {spade, club} (black), then A
and B are mutually exclusive.
Section 7.2 Definition of Probability
The probability of an event occurring is a measure of the proportion of
the time that the event will occur in the long run.
Suppose that in n trials an event E occurs m times. The relative
frequency of the event E is m/n.
If the relative frequency approaches some value P(E) as n becomes
larger, then P(E) is called the empirical probability of E.
Ex. The table below represents the frequency of certain types of license
plates observed by a family on a recent trip. Find the probability
distribution.
State
Number
Probability
Wisconsin
45
45/150 = 0.300
Illinois
80
80/150 = 0.533
Iowa
20
20/150 = 0.133
Indiana
5
5/150 = 0.033
Notice 150 total
observations
Let S = {s1, s2, s3,…,sn} where each si represents a simple event (all
mutually exclusive) and let P(si) represent the probability of event si.
The function P, which assigns a probability to each simple event is
called a probability function.
Also P(si) has the following properties:
0  P( si )  1
probabilities are between 0 and 1
P( s1 )  P( s2 )  ...  P( sn )  1

   P(si )  P(s j )
P si   s j
Sum of the probabilities is 1
Probabilities of the union is the sum
of their probabilities
Probability of an Event in a Uniform Sample Space
If
S = {s1, s2, … , sn}
is the sample space for an experiment in which the outcomes are equally
likely, then we assign the probabilities
P( s1 )  P( s2 )    P( sn ) 
to each of the outcomes s1, s2, … , sn.
1
n
Ex. Assume that when rolling a die each face is equally likely to
show up. If event E = {2} then since S = {1, 2, 3, 4, 5, 6}, we have
P(E) = 1/6. That is, the probability of rolling a 2 is 1 in 6.
Similarly, the probability of rolling any face number is 1/6.
Finding the Probability of an Event E
1. Determine a sample space S associated with the experiment.
2. Assign probabilities to the simple events of S.
3. If E = {s1, s2, s3,…,sn} (each a simple event) then
P(E) = P(s1) + P(s2) +…+ P(sn).
If E is the empty set then P(E) = 0.
Ex. An experiment consists of spinning the hand on the disk below one
time. Assume each outcome is equally likely.
A
C
W
Find P(C) and then find P C  W  .
Notice S = {C, A, W} each of which has a probability of 1/3.
P C  
1
3
P C  W   P(C )  P(W )   
1 1
3 3
2
3
Applied Example: Rolling Dice
• A pair of fair dice is rolled.
• Calculate the probability that the two dice show the same number.
• Calculate the probability that the sum of the numbers of the two dice is
6.
Applied Example 3, page 365
Applied Example: Rolling Dice
Solution
• The sample space S of the experiment has 36 outcomes
S = {(1, 1), (1, 2), … , (6, 5), (6, 6)}
• Both dice are fair, making each of the 36 outcomes equally likely, so we
assign the probability of 1/36 to each simple event.
• The event that the two dice show the same number is
E = {(1, 1), (2, 2) , (3, 3), (4, 4), (5, 5), (6, 6)}
• Therefore, the probability that the two dice show the same number is given
by
P( E )  P[(1,1)]  P[(2, 2)]    P[(6, 6)]
1
1
1 1
    

Six terms
36 36
36 6
Applied Example 3, page 365
Applied Example: Rolling Dice
Solution
• The event that the sum of the numbers of the two dice is 6 is given by
E6 = {(1, 5), (2, 4) , (3, 3), (4, 2), (5, 1)}
• Therefore, the probability that the sum of the numbers on the two dice is 6
is given by
P( E6 )  P[(1,5)]  P[(2, 4)]  P[(3,3)]  P[(4, 2)]  P[(5,1)]

Applied Example 3, page 365
1 1 1 1 1
5
   

36 36 36 36 36 36
Section 7.3 Rules of Probability
Properties of the Probability Function
Property 1
Property 2
P( E )  0 for any E
P( S )  1
If E and F are mutually exclusive (E  F = Ø), then
Property 3
P( E  F )  P( E )  P( F )
Ex. A local grocery store has found kept track of the amount of money spent
by its customers on a single visit. Find the probability that if a customer is
selected at random, the amount spent by the customer will be
a. More than $150
= 0.15
b. More than $50 but less than or equal to $200
Dollars spent
Probability
x  200
0.05
150  x  200
0.10
100  x  150
0.15
50  x  100
x  50
0.25
0.45
= 0.50
Property 4 Addition Rule
If E and F are any two events of an experiment, then
P( E  F )  P( E )  P( F )  P( E  F )
Subtract
overlap
E
F
Note: If E and F are mutually exclusive, then P ( E  F )  0
Ex. A card is drawn from a well-shuffled deck of 52 playing cards. What is
the probability that it is a king or a heart?
K = King and H = Heart
P( K ) 
4
13
1
, P( H )  , P( K  H ) 
52
52
52
P( K  H )  P( K )  P( H )  P( K  H )

4 13 1
 
52 52 52

16 4

52 13
Property 5 Rule of Complements
If E is an event of an experiment and EC denotes the complement of
E, then
P E C  1  P( E )
 
Ex. A card is drawn from a well-shuffled deck of 52 playing cards. What is
the probability that it is not a king?
K = pick a king,
 
P K
C
 1  P( K )  1 
P( K ) 
4
52
48 12
4


52 52 13
Section 7.4 Use of Counting Techniques in Probability
Computing the Probability of an Event in a Uniform Sample Space
Let S be a uniform sample space and let E be any event. Then
number of favorable outcomes in E n  E 
PE 

number of possible outcomes in S
nS 
Ex. Suppose that you reach into a box of 12 size AA batteries and you
know that 4 of them are dead. Find the probability that
a. in one draw you get a good battery.
n  good batteries 
n  batteries 

C  8,1
C 12,1

8 2

12 3
b. in two draws without replacement you get two good batteries.
n  ways to get 2 good 
n  ways to draw 2 batteries 

C  8, 2 
C 12, 2 

28 14

66 33
Ex. Three balls are selected at random without replacement from
the jar below. Find the probability that
a. All 3 of the balls are green.
n  draw 3 green 
n  draw 3

C  3,3
C  8,3

1
56
b. One ball is red and two are black.
n  draw 1 red, 2 black 
n  draw 3

C  2,1  C  3, 2 
C  8,3
6
3


56 28
Ex. Refer to the jar of marbles below. Two marbles are drawn at
random without replacement.
Find the probability that no yellow are drawn.
1  P  both yellow   1 
C  3, 2 
C 11, 2 
 1
3 52

55 55
Section 7.5 Conditional Probability and Independent Events
The probability of an event is affected by the knowledge of other
information relevant to the event.
Notation: P(A|B) is read “the probability of event A given that event B
has occurred.”
Ex. You roll a fair die. Find the probability that you roll a 2 given that
your roll is an even.
Knowing it is even restricts the sample space to {2, 4, 6}.
So
P  2 | even   1
3
Conditional Probability of an Event
If A and B are events in an experiment and P( A)  0, then the conditional
probability that the event B will occur given that A has already occurred is
P  B | A 
P  A  B
P  A
Which can be written (the Product Rule):
P  A  B   P  A  P  B | A 
Ex. In a box of 20 size AA batteries, 10 are brand X and 10 are brand Y.
You also know that 3 of the brand X batteries are dead, while 2 of the brand
Y are dead. Find the probability that in a (random) draw
a. you get a dead (D) brand X battery.
P  X  D  P  X   P  D | X 

10 3
3


20 10
20
b. you get brand Y given that you drew a dead (D) battery.
P Y | D  
P Y  D 
P  D

2 / 20
2

5 / 20
5
Independent Events
If A and B are independent events, then P  B | A  P( B)
P( A | B)  P( A)
Test for Independent Events
Events A and B are independent events if and only if P  A  B   P( A) P( B)
Note: this generalizes to more than two independent events.
Ex. If A die is rolled twice, show that rolling a 5 on the first roll and rolling
a 4 on the second roll are independent events.
(roll 1, roll2)
 1,1

(2,1)
 (3,1)
S 
(4,1)
 (5,1)

(6,1)
(1, 2)
(1,3)
(1, 4)
(1,5)
(2, 2) (2,3) (2, 4) (2,5)
(3, 2) (3,3) (3, 4) (3,5)
(4, 2) (4,3) (4, 4) (4,5)
(5, 2) (5,3) (5, 4) (5,5)
(6, 2) (6,3) (6, 4) (6,5)
V = 5 on first roll, R = 4 on second roll
1
P V  R  
36
P V   P  R 

Therefore V and R are independent
6 6
1
 
36 36 36
(1, 6) 

(2, 6) 
(3, 6) 

(4, 6) 
(5, 6) 

(6, 6) 
Section 7.6 Bayes’ Theorem
Bayes’ Theorem
Let A1, A2, …, An be a partition of a sample space S and let E be an event of
the experiment such that P(E) is not zero. Then the posteriori probability
P(Ai|E) is given by
P  Ai | E  
P  Ai  P  E | Ai 
P  A1  P  E | A1   P  A2  P  E | A2   ...  P  An  P  E | An 
Where
1 i  n
Posteriori probability: probability is calculated after the outcomes of the
experiment have occurred.
Ex. A store stocks light bulbs from three suppliers. Suppliers A, B, and C
supply 10%, 20%, and 70% of the bulbs respectively. It has been determined
that company A’s bulbs are 1% defective while company B’s are 3% defective
and company C’s are 4% defective. If a bulb is selected at random and found
to be defective, what is the probability that it came from supplier B?
Let D = defective
P  B | D 

P  B P  D | B
P  A P  D | A  P  B  P  D | B   P C  P  D | C 
0.2  0.03
0.1 0.01  0.2  0.03  0.7  0.04 
So about 0.17
 0.1714