# Probability by fjzhangxiaoquan

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```									Math I
Probability Experiments
Probability experiment
• An action, or trial, through which specific results (counts,
measurements, or responses) are obtained.
Outcome
• The result of a single trial in a probability experiment.
Sample Space
• The set of all possible outcomes of a probability
experiment.
Event
• Consists of one or more outcomes and is a subset of the
sample space.
Larson/Farber 4th ed                                                3
Probability Experiments

• Probability experiment: Roll a die

• Outcome: {3}

• Sample space: {1, 2, 3, 4, 5, 6}

• Event: {Die is even}={2, 4, 6}

Larson/Farber 4th ed                             4
Example: Identifying the Sample Space

A probability experiment consists of tossing a coin and
then rolling a six-sided die. Describe the sample space.

Solution:
There are two possible outcomes when tossing a coin:
a head (H) or a tail (T). For each of these, there are six
possible outcomes when rolling a die: 1, 2, 3, 4, 5, or
6. One way to list outcomes for actions occurring in a
sequence is to use a tree diagram.

Larson/Farber 4th ed                                              5
Solution: Identifying the Sample Space

Tree diagram:

H1 H2 H3 H4 H5 H6    T1 T2 T3 T4 T5 T6

The sample space has 12 outcomes:
{H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}

Larson/Farber 4th ed                                     6
Simple Events

Simple event
• An event that consists of a single outcome.
 e.g. “Tossing heads and rolling a 3” {H3}

• An event that consists of more than one outcome is
not a simple event.
 e.g. “Tossing heads and rolling an even number”
{H2, H4, H6}

Larson/Farber 4th ed                                      7
Example: Identifying Simple Events

Determine whether the event is simple or not.
• You roll a six-sided die. Event B is rolling at least a 4.

Solution:
Not simple (event B has three outcomes: rolling a 4, a 5,
or a 6)

Larson/Farber 4th ed                                              8
   Probability is the chance that something will happen.

   Probability is most often expressed as a fraction, a
decimal, a percent, or can also be written out in words.

   To determine the probability…

P(event) =         number of true outcomes
total number of equally likely outcomes
   Independent: Two events are independent if
the occurrence of one has no effect on the
occurrence of the other…
   Example 1: (probability of two events)
What is the probability of drawing a king and then
an ace from a standard 52 card deck with
replacement?                      4 4        16
P(King, Ace) =       
52 52     2704
   Example 2:
What is the probability of flipping heads on a coin
three times in a row?                1 1 1 1
P(H, H, H) =   
2 2 2      8
   Example 3:
A die is rolled twice. What’s the probability of rolling a 2
and then an even number?
1 1 1
Solution:   
6 2 12
   Example 4:
You spin the spinner 3 times.
What is the probability of spinning a 4, a 3 and then a 1?

Solution:      1 2 3  6   3
     
8 8 8 512 256
Complementary Events

Complement of event E
• The set of all outcomes in a sample space that are not
included in event E.
• Denoted E ′ (E prime)
• P(E ′) + P(E) = 1
• P(E) = 1 – P(E ′)                               E′
• P(E ′) = 1 – P(E)                    E

Larson/Farber 4th ed                                          13
Let’s try a few
Complement
Spinner
• In the spinner
at the right:
• P’(yellow)
• P’(not blue)
• P’(red or yellow)
• P’(white)
• P’(odd number)
   Dependent: Two events such that the
occurrence of one affects the occurrence of the


other.
P(A and B) = P(A) P(B|A)

**P(B|A) = means the probability of B
given that event A has already
occurred.
   Dependent: WITHOUT REPLACEMENT

   Independent: WITH REPLACEMENT
Independent and Dependent Events

Independent events
• The occurrence of one of the events does not affect
the probability of the occurrence of the other event
• P(B | A) = P(B) or P(A | B) = P(A)
• Events that are not independent are dependent

Larson/Farber 4th ed                                        18
Example: Independent and Dependent
Events
Decide whether the events are independent or dependent.
1. Selecting a king from a standard deck (A), not
replacing it, and then selecting a queen from the deck
(B).

Solution:
4
P( B | A)  P(2nd card is a Queen |1st card is a King ) 
51
4
P( B)  P(Queen) 
52
Dependent (the occurrence of A changes the probability
of the occurrence of B)
Larson/Farber 4th ed                                                     19
Example: Independent and Dependent
Events
Decide whether the events are independent or dependent.
2. Tossing a coin and getting a head (A), and then
rolling a six-sided die and obtaining a 6 (B).
Solution:
1
P( B | A)  P(rolling a 6 | head on coin) 
6
1
P( B)  P(rolling a 6) 
6

Independent (the occurrence of A does not change the
probability of the occurrence of B)
Larson/Farber 4th ed                                      20
   You Spin A Spinner Twice
◦ Independent
   You choose a pair of shoes and a pair of
socks
◦ Independent
   You buy two different CD’s
◦ Dependent
   You roll a number cube twice so that6 the
sum is 8
◦ Dependent
   You spin a spinner and flip a coin
◦ Independent
   You pick a marble and pick again without
replacing.
◦ Dependent
Example: Using the Multiplication Rule

Two cards are selected, without replacing the first card,
from a standard deck. Find the probability of selecting a
king and then selecting a queen.
Solution:
Because the first card is not replaced, the events are
dependent.
P( K and Q)  P ( K )  P (Q | K )
4 4
 
52 51
16
       0.006
2652
Larson/Farber 4th ed                                           23
Example: Using the Multiplication Rule

A coin is tossed and a die is rolled. Find the probability
of getting a head and then rolling a 6.
Solution:
The outcome of the coin does not affect the probability
of rolling a 6 on the die. These two events are
independent.
P ( H and 6)  P ( H )  P (6)
1 1
    
2 6
1
       0.083
12
Larson/Farber 4th ed                                            24
Example: Using the Multiplication Rule

The probability that a particular knee surgery is
successful is 0.85. Find the probability that three knee
surgeries are successful.

Solution:
The probability that each knee surgery is successful is
0.85. The chance for success for one surgery is
independent of the chances for the other surgeries.
P(3 surgeries are successful) = (0.85)(0.85)(0.85)
≈ 0.614
Larson/Farber 4th ed                                          25
Example: Using the Multiplication Rule

Find the probability that none of the three knee
surgeries is successful.

Solution:
Because the probability of success for one surgery is
0.85. The probability of failure for one surgery is
1 – 0.85 = 0.15
P(none of the 3 surgeries is successful) = (0.15)(0.15)(0.15)
≈ 0.003

Larson/Farber 4th ed                                              26
Example: Using the Multiplication Rule

Find the probability that at least one of the three knee
surgeries is successful.

Solution:
“At least one” means one or more. The complement to
the event “at least one successful” is the event “none are
successful.” Using the complement rule
P(at least 1 is successful) = 1 – P(none are successful)
≈ 1 – 0.003
= 0.997
Larson/Farber 4th ed                                            27
Example: Using the Multiplication Rule to
Find Probabilities
More than 15,000 U.S. medical school seniors applied to
residency programs in 2007. Of those, 93% were matched to
a residency position. Seventy-four percent of the seniors
matched to a residency position were matched to one of their
top two choices. Medical students electronically rank the
residency programs in their order of preference and program
directors across the United States do the same. The term
“match” refers to the process where a student’s preference
list and a program director’s preference list overlap,
resulting in the placement of the student for a residency
position. (Source: National Resident Matching Program)
(continued)
Larson/Farber 4th ed                                              28
Example: Using the Multiplication Rule to
Find Probabilities
1. Find the probability that a randomly selected senior was
matched a residency position and it was one of the
senior’s top two choices.
Solution:
A = {matched to residency position}
B = {matched to one of two top choices}
P(A) = 0.93 and P(B | A) = 0.74
P(A and B) = P(A)∙P(B | A) = (0.93)(0.74) ≈ 0.688
dependent events

Larson/Farber 4th ed                                             29
Example: Using the Multiplication Rule to
Find Probabilities
2. Find the probability that a randomly selected senior that
was matched to a residency position did not get matched
with one of the senior’s top two choices.
Solution:
Use the complement:
P(B′ | A) = 1 – P(B | A)
= 1 – 0.74 = 0.26

Larson/Farber 4th ed                                              30
   Example 1:
What is the probability of drawing a King
and then an Ace without replacement?
4 4        16
P(King, Ace) = 52  51  2652
   Example 2:
You randomly select two marbles from a bag that
contains 14 green, 7 blue, and 9 red marbles. What
is the probability that the first marble is blue and
the second marble is not blue if you do not replace
the first marble?

Solution:
7 23 161
  
30 29 870
   Example 3:
erasers, 9 blue erasers, and 7 green erasers. If you
and your two neighbors are the first to randomly
select an eraser, what is the probability that all
three of you select green erasers?

Solution:
P(A) and P(B|A) and P(C|A and B)           7 6 5     210
    
22 21 20 9240
   The table shows the number of males and
females with certain hair colors. Find …

 A) the probability that a listed person has red hair
 B) the probability that a female has red hair

Brown    Blonde Red           Black     Other
hair     hair   hair          hair
Male       42       11     3             17        27
Female 47           16         13        9         15
   P(red hair) = # of people with red hair 16   2

total # of people        200 25

P(red hair | female) = # of red hair females      13


total # of females       100
 When  you consider the outcomes for
either of two events A and B, you form
the union of A and B.
   When you consider only the outcomes shared
by both A and B, you form the intersection of
A and B.
   When the sets of A and B have nothing in
common (no intersection) then they are
considered mutually exclusive events.
Mutually exclusive
 Two events A and B cannot occur at the same time

A
B               A              B

A and B are               A and B are not
mutually                  mutually exclusive
exclusive
40   Larson/Farber 4th ed
Decide if the events are mutually exclusive.
Event A: Roll a 3 on a die.
Event B: Roll a 4 on a die.

Solution:
Mutually exclusive (The first event has one
outcome, a 3. The second event also has one
outcome, a 4. These outcomes cannot occur
at the same time.)

41   Larson/Farber 4th ed
Decide if the events are mutually exclusive.
Event A: Randomly select a male student.
Event B: Randomly select a nursing major.

Solution:
Not mutually exclusive (The student can be a
male nursing major.)

42   Larson/Farber 4th ed
   If A and B are mutually exclusive events (one
event does not have anything in common with
the other), then…

P(A or B) = P(A) + P(B)
   A die is rolled one time. What is the
probability of rolling a 2 or a 6?

Solution:                                     1

1 1 2
P(A or B) = P(A) + P(B) =   
6   6     6      3
   A card is randomly selected out of a
standard deck of 52 cards. What is the
probability that it is a 2 or a king?

P(A or B) = P(A) + P(B) =
2
4

52 52
4

8
52

13
 If A and B are not mutually exclusive, then
there are some outcomes in common.
 Therefore, the intersection of A and B are
counted twice when P(A) and P(B) are added.
 So, P(A and B) must be subtracted once from
the sum…
P(A or B) = P(A) + P(B) – P(A and B)
 A die is rolled one time. What is the
probability of rolling an odd number or
a prime number?
3
 Odd = 1, 3, 5     P(A) = 6
Prime = 2, 3, 5 P(B) = 3
6
2
 Odd and prime = 3, 5 P(A and B) =
6
 A card is randomly selected from standard
deck of 52 cards. What is the probability
that it is a red card or a king?
 Red cards = 26 =     26
52
Kings = 4 =   4
52
2
   Red Kings = 2 =   52
   The probability that it will rain today is 40%. The
probability that is will rain tomorrow is 30%. The
probability that it will rain both days is 20%. Find
the probability that it will rain either today OR
tomorrow.
   Solution: P(A) + P(B) – P(A and B)

P(today) + P(tomorrow) – P(today and tomorrow)
=
40%  30%  20%  50%

Addition rule for the probability of A or B
• The probability that events A or B will occur is
 P(A or B) = P(A) + P(B) – P(A and B)
• For mutually exclusive events A and B, the rule can
be simplified to
 P(A or B) = P(A) + P(B)
 Can be extended to any number of mutually
exclusive events

Larson/Farber 4th ed                                       50

You select a card from a standard deck. Find the
probability that the card is a 4 or an ace.

Solution:
The events are mutually exclusive (if the card is a 4, it
cannot be an ace)                     Deck of 52 Cards
P (4 or ace)  P (4)  P( ace)
4♣
4   4                        4♥
                    4♠               A♣
52 52                    4♦        A♠ A♥
8                                  A♦
     0.154      44 other cards
52
Larson/Farber 4th ed                                                51

You roll a die. Find the probability of rolling a number
less than 3 or rolling an odd number.

Solution:
The events are not mutually exclusive (1 is an
outcome of both events)             Roll a Die
4     6
Odd    Less than
3         1 three
5         2

Larson/Farber 4th ed                                          52
Roll a Die
4      6
Odd     Less than
3          1 three
5           2

P(less than 3 or odd )
 P(less than 3)  P(odd )  P(less than 3 and odd )
2 3 1 4
     0.667
6 6 6 6
Larson/Farber 4th ed                                             53

The frequency distribution shows     Sales volume (\$)   Months
the volume of sales (in dollars)            0–24,999      3
and the number of months a sales       25,000–49,999      5
representative reached each sales      50,000–74,999      6
level during the past three years.     75,000–99,999      7
If this sales pattern continues,     100,000–124,999      9
what is the probability that the     125,000–149,999      2
sales representative will sell       150,000–174,999      3
between \$75,000 and \$124,999         175,000–199,999      1

next month?

Larson/Farber 4th ed                                                54

• A = monthly sales between                Sales volume (\$)   Months
\$75,000 and \$99,999                             0–24,999      3
• B = monthly sales between                  25,000–49,999      5
\$100,000 and \$124,999                      50,000–74,999      6

• A and B are mutually exclusive             75,000–99,999      7
100,000–124,999      9
P ( A or B )  P ( A)  P ( B )   125,000–149,999      2
7   9             150,000–174,999      3
    
36 36              175,000–199,999      1
16
     0.444
36
Larson/Farber 4th ed                                                      55

A blood bank catalogs the types of blood given by
donors during the last five days. A donor is selected at
random. Find the probability the donor has type O or
type A blood.

Type O   Type A   Type B   Type AB   Total
Rh-Positive      156      139      37        12      344
Rh-Negative       28       25       8         4       65
Total            184      164      45        16      409

Larson/Farber 4th ed                                                56

The events are mutually exclusive (a donor cannot have
type O blood and type A blood)
Type O    Type A   Type B   Type AB    Total
Rh-Positive       156      139       37        12      344
Rh-Negative        28       25        8        4        65
Total             184      164       45        16      409

P (type O or type A)  P (type O )  P (type A)
184 164
     
409 409
348
      0.851
409
Larson/Farber 4th ed                                                    57

Find the probability the donor has type B or is Rh-
negative.
Type O   Type A   Type B   Type AB   Total
Rh-Positive    156      139      37        12      344
Rh-Negative     28       25       8         4       65
Total          184      164      45        16      409

Solution:
The events are not mutually exclusive (a donor can have
type B blood and be Rh-negative)

Larson/Farber 4th ed                                                58

Type O   Type A   Type B   Type AB   Total
Rh-Positive    156      139      37        12      344
Rh-Negative     28       25       8         4       65
Total          184      164      45        16      409

P(type B or Rh  neg )
 P(type B)  P( Rh  neg )  P(type B and Rh  neg )
45    65     8    102
                        0.249
409 409 409 409

Larson/Farber 4th ed                                                59
   Formula

   Means the probability of B given that A has
◦ The probability of A and B occurring
 Divided by the probability of A

   Example 1:
◦ Gus rolled a six-sided cube, with faces numbered 1 to 6,
twice. If you know that he rolled a 2 on his first roll,
what is the probability that he rolled two even numbers?
 Find the probability of getting two even numbers given that
he rolled a two
 Probability of A and B is 1/6 times ½ which equals 1/12
 Divide that by 1/6
 = 1/2
   Example 2:
◦ Antoine will toss a fair coin three times. What is the
probability that all 3 tosses will result in heads if you
know that his first toss resulted in heads?
 Find the probability of getting all three heads given that the
 Probability of getting all three heads is ½ times ½ times ½
which is 1/8
 Divide that by 1/2
 = 1/4
   Example 3:

 A bag contains 4 red marbles, 3 blue marbles, and 2
green marbles. What is the probability that Nick reached
into the bag and picked a red marble if you know that he did
NOT pick a green marble?
 Did not say given that
 Don’t Over Think…..If he did not pick green, that means there
were only 7 marbles left. His probability was 4/7.
   Example 4:
◦ Calculate the probability of rolling a total of 8 if the first
roll is a 5.
 P(5) = 1/6 P(sum 8) = 3/6 P(A and B)= 3/36
 Divide by 1/6
 = 1/6
   The expected value is often referred to as the
“long-term” average or mean .

   This means that over the long term of doing an
experiment over and over, you would expect this
average.

   To find the expected value or long term average,
simply multiply each value of the random
variable by its probability and add the products.
   Suppose that the following game is played. A
man rolls a die. If he rolls a 1, 3, or 5, he loses
\$3, if he rolls a 4 or 6, he loses \$2, and if he
rolls a 2, he wins \$12. What gains or losses
should he expect on average? (What is his
expected value?)
   We must find the probabilities of each outcome. We
can make a chart to help us see this.

Possible        Probability
Losses/Gains    P(losses/gains)

-\$3               3
6

-\$2                2
6

\$12
1
6
   Now, we multiply the probability for each outcome by
the amount of money either gained or lost for that
outcome.
3    9
 \$3     \$1.50
      Expected value of rolling a 1, 3, or 5: ________________
6    6
2    4
 \$2     \$0.67
      Expected value of rolling a 4 or 6:    ________________
6    6

 1  12
\$12        \$2.00
      Expected value of rolling a 2:         _______________
6 6
   To find the expected value for the entire game