probability - anne gloag's math page...
TRANSCRIPT
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Chapter
Probability
3
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Section 3.1 Basic Concepts of Probability
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Identify the sample space of a probability experiment
Identify simple events
Use the Fundamental Counting Principle
Distinguish among classical probability, empirical probability, and subjective probability
Determine the probability of the complement of an event
Use a tree diagram and the Fundamental Counting Principle to find probabilities
Section 3.1 Objectives
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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.
Probability Experiments
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Probability experiment: Roll a die
Outcome: {3}
Sample space: {1, 2, 3, 4, 5, 6}
Event: {Die is even}={2, 4, 6}
Probability Experiments
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A probability experiment consists of tossing a coin and then rolling a six-sided die. Describe the sample space.
Example: Identifying 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.
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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}
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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}
Simple Events
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Fundamental Counting Principle
If one event can occur in m ways and a second event can occur in n ways, the number of ways the two events can occur in sequence is m•n.
Can be extended for any number of events occurring in sequence.
Fundamental Counting Principle
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You are purchasing a new car. The possible manufacturers, car sizes, and colors are listed.
Manufacturer: Ford, GM, Honda
Car size: compact, midsize
Color: white (W), red (R), black (B), green (G)
How many different ways can you select one manufacturer, one car size, and one color? Use a tree diagram to check your result.
Example: Fundamental Counting Principle
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There are three choices of manufacturers, two car sizes, and four colors.
Using the Fundamental Counting Principle:
3 ∙ 2 ∙ 4 = 24 ways
Solution: Fundamental Counting Principle
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In a particular college class, there are male and female students. Some students have long hair and some students have short hair. Write the symbols for the probabilities of the events for parts (a) through (j) below. (Note that you can't find numerical answers here. You were not given enough information to find any probability values yet; concentrate on understanding the symbols.)
Let F be the event that a student is female.
Let M be the event that a student is male.
Let S be the event that a student has short hair.
Let L be the event that a student has long hair.
Exercise
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a. The probability that a student does not have long hair. b. The probability that a student is male or has short hair. c. The probability that a student is a female and has long hair. d. The probability that a student is male, given that the student has long hair. e. The probability that a student has long hair, given that the student is male. f. Of all the female students, the probability that a student has short hair. g. Of all students with long hair, the probability that a student is female. h. The probability that a student is female or has long hair. i. The probability that a randomly selected student is a male student with short hair. j. The probability that a student is female.
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a. P(L')=P(S) b. P(M or S) c. P(F and L) d. P(M|L) e. P(L|M) f. P(S|F) g. P(F|L) h. P(F or L) i. P(M and S) j. P(F)
Solutions
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Classical (theoretical) Probability
Each outcome in a sample space is equally likely.
Types of Probability
Number of outcomes in event E( )
Number of outcomes in sample spaceP E
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1. Event A: rolling a 3
2. Event B: rolling a 7
3. Event C: rolling a number less than 5
Example: Finding Classical Probabilities
Solution: Sample space: {1, 2, 3, 4, 5, 6}
You roll a six-sided die. Find the probability of each
event.
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1. Event A: rolling a 3 Event A = {3}
Solution: Finding Classical Probabilities
1( 3) 0.167
6P rolling a
2. Event B: rolling a 7 Event B= { } (7 is not in
the sample space) 0( 7) 0
6P rolling a
3. Event C: rolling a number less than 5
Event C = {1, 2, 3, 4} 4
( 5) 0.6676
P rolling a number less than
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Empirical (statistical) Probability
Based on observations obtained from probability experiments.
Relative frequency of an event.
Types of Probability
Frequency of event E( )
Total frequency
fP E
n
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1. A company is conducting a telephone survey of randomly selected individuals to get their overall impressions of the past decade (2000s). So far, 1504 people have been surveyed. What is the probability that the next person surveyed has a positive overall impression of the 2000s? (Source: Princeton Survey Research Associates International)
Example: Finding Empirical Probabilities
Response Number of times, f
Positive 406
Negative 752
Neither 316
Don’t know 30
Σf = 1504
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Solution: Finding Empirical Probabilities
Response Number of times, f
Positive 406
Negative 752
Neither 316
Don’t know 30
Σf = 320
event frequency
406( ) 0.270
1504
fP positive
n
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Law of Large Numbers
As an experiment is repeated over and over, the empirical probability of an event approaches the theoretical (actual) probability of the event.
http://www.statcrunch.com/app/index.php?dataid=205337
Law of Large Numbers
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Range of probabilities rule
The probability of an event E is between 0 and 1, inclusive.
0 ≤ P(E) ≤ 1
Range of Probabilities Rule
[ ] 0 0.5 1
Impossible Unlikely Even
chance Likely Certain
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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 ′)
P(E ′) = 1 – P(E)
Complementary Events
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You survey a sample of 1000 employees at a company and record the age of each. Find the probability of randomly choosing an employee who is not between 25 and 34 years old.
Example: Probability of the Complement of an Event
Employee ages Frequency, f
15 to 24 54
25 to 34 366
35 to 44 233
45 to 54 180
55 to 64 125
65 and over 42
Σf = 1000
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Use empirical probability to find P(age 25 to 34)
Solution: Probability of the Complement of an Event
Employee ages Frequency, f
15 to 24 54
25 to 34 366
35 to 44 233
45 to 54 180
55 to 64 125
65 and over 42
Σf = 1000
P(age 25 to 34)
f
n
366
1000 0.366
• Use the complement rule
P(age is not 25 to 34) 1366
1000
634
1000 0.634
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A probability experiment consists of tossing a coin and spinning the spinner shown. The spinner is equally likely to land on each number. Use a tree diagram to find the probability of tossing a tail and spinning an odd number.
Example: Probability Using a Tree Diagram
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Tree Diagram:
Solution: Probability Using a Tree Diagram
H T
1 2 3 4 5 7 6 8 1 2 3 4 5 7 6 8
H1 H2 H3 H4 H5 H6 H7 H8 T1 T2 T3 T4 T5 T6 T7 T8
P(tossing a tail and spinning an odd number) = 4 1
0.2516 4
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Your college identification number consists of 8 digits. Each digit can be 0 through 9 and each digit can be repeated. What is the probability of getting your college identification number when randomly generating eight digits?
Example: Probability Using the Fundamental Counting Principle
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Each digit can be repeated
There are 10 choices for each of the 8 digits
Using the Fundamental Counting Principle, there are 10 ∙ 10 ∙ 10 ∙ 10 ∙ 10 ∙ 10 ∙ 10 ∙ 10
= 108 = 100,000,000 possible identification numbers
Only one of those numbers corresponds to your ID number
Solution: Probability Using the Fundamental Counting Principle
1
100,000,000P(your ID number) =
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Section 3.2
Conditional Probability and the Multiplication Rule
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Determine conditional probabilities
Distinguish between independent and dependent events
Use the Multiplication Rule to find the probability of two events occurring in sequence
Use the Multiplication Rule to find conditional probabilities
Section 3.2 Objectives
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Conditional Probability
The probability of an event occurring, given that another event has already occurred
Denoted P(B | A) (read “probability of B, given A”)
Conditional Probability
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Two cards are selected in sequence from a standard deck. Find the probability that the second card is a queen, given that the first card is a king. (Assume that the king is not replaced.)
Example: Finding Conditional Probabilities
Solution:
Because the first card is a king and is not replaced, the
remaining deck has 51 cards, 4 of which are queens.
4
( | ) (2 |1 ) 0.07851
nd stP B A P card is a Queen card is a King
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The table shows the results of a study in which researchers examined a child’s IQ and the presence of a specific gene in the child. Find the probability that a child has a high IQ, given that the child has the gene.
Example: Finding Conditional Probabilities
Gene
Present
Gene not
present
Total
High IQ 33 19 52
Normal IQ 39 11 50
Total 72 30 102
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There are 72 children who have the gene. So, the sample space consists of these 72 children.
Solution: Finding Conditional Probabilities
P(B | A) P(high IQ | gene present )
33
72 0.458
Of these, 33 have a high IQ.
Gene
Present
Gene not
present
Total
High IQ 33 19 52
Normal IQ 39 11 50
Total 72 30 102
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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
Independent and Dependent Events
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1. Selecting a king from a standard deck (A), not replacing it, and then selecting a queen from the deck (B).
Example: Independent and Dependent Events
4( | ) (2 |1 )
51
nd stP B A P card is a Queen card is a King
4( ) ( )
52P B P Queen
Dependent (the occurrence of A changes the probability of the occurrence of B)
Solution:
Decide whether the events are independent or dependent.
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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).
Example: Independent and Dependent Events
1( | ) ( 6 | )
6P B A P rolling a head on coin
1( ) ( 6)
6P B P rolling a
Independent (the occurrence of A does not change the probability of the occurrence of B)
Solution:
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Multiplication rule for the probability of A and B
The probability that two events A and B will occur in sequence is
P(A and B) = P(A) ∙ P(B | A)
For independent events the rule can be simplified to
P(A and B) = P(A) ∙ P(B)
Can be extended for any number of independent events
The Multiplication Rule
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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.
Example: Using the Multiplication Rule
Solution: Because the first card is not replaced, the events are dependent.
( ) ( ) ( | )
4 4
52 51
160.006
2652
P K and Q P K P Q K
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A coin is tossed and a die is rolled. Find the probability of getting a head and then rolling a 6.
Example: Using the Multiplication Rule
Solution: The outcome of the coin does not affect the probability of rolling a 6 on the die. These two events are independent. ( 6) ( ) (6)
1 1
2 6
10.083
12
P H and P H P
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The probability that a particular knee surgery is successful is 0.85. Find the probability that three knee surgeries are successful.
Example: Using the Multiplication Rule
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
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Find the probability that none of the three knee surgeries is successful.
Example: Using the Multiplication Rule
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
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Find the probability that at least one of the three knee surgeries is successful.
Example: Using the Multiplication Rule
Solution:
“At least one” means one or more. The complement to
the event “at least one is 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
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Two events are independent if the following are true:
P(A|B) = P(A)
P(B|A) = P(B)
P(A AND B) = P(A) ⋅ P(B)
Two events A and B are independent if the knowledge that one occurred does not affect the chance the other occurs.
For example, the outcomes of two roles of a fair die are independent events. The outcome of the first roll does not change the probability for the outcome of the second roll. To show two events are independent, you must show only one of the above conditions. If two events are NOT independent, then we say that they are dependent.
More on Independent Events
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Sampling may be done with replacement or without replacement. With replacement: If each member of a population is replaced
after it is picked, then that member has the possibility of being chosen more than once. When sampling is done with replacement, then events are considered to be independent, meaning the result of the first pick will not change the probabilities for the second pick.
Without replacement: When sampling is done without replacement, then each member of a population may be chosen only once. In this case, the probabilities for the second pick are affected by the result of the first pick. The events are considered to be dependent or not independent.
If it is not known whether A and B are independent or dependent, assume they are dependent until you can show otherwise
With and without replacement
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Section 3.3
Addition Rule
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Determine if two events are mutually exclusive
Use the Addition Rule to find the probability of two events
Section 3.3 Objectives
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Decide if the events are mutually exclusive.
Event A: Roll a 3 on a die.
Event B: Roll a 4 on a die.
Example: Mutually Exclusive Events
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.)
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Decide if the events are mutually exclusive.
Event A: Randomly select a male student.
Event B: Randomly select a nursing major.
Example: Mutually Exclusive Events
Solution:
Not mutually exclusive (The student can be a male
nursing major.)
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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
The Addition Rule
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You select a card from a standard deck. Find the probability that the card is a 4 or an ace.
Example: Using the Addition Rule
Solution: The events are mutually exclusive (if the card is a 4, it cannot be an ace)
(4 ) (4) ( )
4 4
52 52
80.154
52
P or ace P P ace
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You roll a die. Find the probability of rolling a number less than 3 or rolling an odd number.
Example: Using the Addition Rule
Solution: The events are not mutually exclusive (1 is an outcome of both events)
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Solution: Using the Addition Rule
( 3 )
( 3) ( ) ( 3 )
2 3 1 40.667
6 6 6 6
P less than or odd
P less than P odd P less than and odd
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The frequency distribution shows the volume of sales (in dollars) and the number of months in which a sales representative reached each sales level during the past three years. If this sales pattern continues, what is the probability that the sales representative will sell between $75,000 and $124,999 next month?
Example: Using the Addition Rule
Sales volume ($) Months
0–24,999 3
25,000–49,999 5
50,000–74,999 6
75,000–99,999 7
100,000–124,999 9
125,000–149,999 2
150,000–174,999 3
175,000–199,999 1
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A = monthly sales between $75,000 and $99,999
B = monthly sales between $100,000 and $124,999
A and B are mutually exclusive
Solution: Using the Addition Rule
Sales volume ($) Months
0–24,999 3
25,000–49,999 5
50,000–74,999 6
75,000–99,999 7
100,000–124,999 9
125,000–149,999 2
150,000–174,999 3
175,000–199,999 1
( ) ( ) ( )
7 9
36 36
160.444
36
P A or B P A P B
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A blood bank catalogs the types of blood, including positive or negative Rh-factor, given by donors during the last five days. A donor is selected at random. Find the probability that the donor has type O or type A blood.
Example: Using the Addition Rule
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
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Solution: Using the Addition Rule
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
409
164
409
348
409 0.851
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Find the probability that the donor has type B blood or is Rh-negative.
Example: Using the Addition Rule
Solution:
The events are not mutually exclusive (a donor can have
type B blood and be 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
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Solution: Using the Addition Rule
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
409
65
409
8
409
102
409 0.249
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A tree diagram is a special type of graph used to determine the outcomes of an experiment. It consists of "branches" that are labeled with either frequencies or probabilities. Tree diagrams can make some probability problems easier to visualize and solve. The following example illustrates how to use a tree diagram.
Tree Diagrams
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In an urn, there are 11 balls. Three balls are red (R) and 8 balls are blue (B). Draw two balls, one at a time, with replacement. "With replacement" means that you put the first ball back in the urn before you select the second ball. The tree diagram using frequencies that show all the possible outcomes follows.
Larson/Farber 5th edition 62
EXAMPLE
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1. List the 24 BR outcomes: B1R1, B1R2, B1R3, ...
2. Using the tree diagram, calculate P(RR).
3. Using the tree diagram, calculate P(RB OR BR).
4. Using the tree diagram, calculate P(R on 1st draw AND B on 2nd draw).
5. Using the tree diagram, calculate P(R on 2nd draw given B on 1st draw).
6. Using the tree diagram, calculate P(BB).
7. Using the tree diagram, calculate P(B on the 2nd draw given R on the first draw).
Larson/Farber 5th edition 64
Calculate
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1. B1R1; B1R2; B1R3; B2R1; B2R2; B2R3; B3R1; B3R2; B3R3; B4R1; B4R2; B4R3; B5R1; B5R2; B5R3; B6R1; B6R2; B6R3; B7R1; B7R2; B7R3; B8R1; B8R2; B8R3 2. P(RR)=(3/11)⋅(3/11)=9/121 3. P(RB OR BR)=(3/11)⋅(8/11) + (8/11)⋅(3/11) = 48/121 4. P(R on 1st draw AND B on 2nd draw)=P(RB)=(3/11)⋅(8/11)=24/121 5. P(R on 2nd draw given B on 1st draw)=P(R on 2nd | B on 1st)=24/88=3/11 6. P(BB) = 64/121 7. P(B on 2nd draw | R on 1st draw) = 8/11 There are 9 + 24 outcomes that have R on the first draw (9 RR and 24 RB).
Answers
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An urn has 3 red marbles and 8 blue marbles in it. Draw two marbles, one at a time, this time without replacement from the urn. "Without replacement" means that you do not put the first ball back before you select the second ball. Below is a tree diagram. The branches are labeled with probabilities instead of frequencies. The numbers at the ends of the branches are calculated by multiplying the numbers on the two corresponding branches, for example, (3/11)⋅(2/10)=6/110.
Example
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1. P(RR) =
2. P(RB OR BR)=
3. P(R on 2d | B on 1st) =
4. P(R on 1st and B on 2nd) =
5. P(BB) =
6. P(B on 2nd | R on 1st) =
Calculate
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1. P(RR)=(3/11)⋅(2/10)=6/110
2. P(RB or BR)=(3/11)⋅(8/10) + (8/11)(3/10) =48/110
3. P(R on 2d | B on 1st) = 310
4. P(R on 1st and B on 2nd) = P(RB) = (3/11)(8/10) = 24/110
5. P(BB) = (8/11) ⋅ (7/10)
6. The probability is 24/30
Answers