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![Page 1: The Practice of Statistics Third Edition Chapter 6: Probability and Simulation: The Study of Randomness 6.3 General Probability Rules Copyright © 2008.](https://reader035.fdocuments.in/reader035/viewer/2022062321/56649dbd5503460f94aaf573/html5/thumbnails/1.jpg)
The Practice of StatisticsThird Edition
Chapter 6:Probability and Simulation:The Study of Randomness
6.3 General Probability Rules
Copyright © 2008 by W. H. Freeman & Company
Daniel S. Yates
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Essential Questions
• What is the addition rule for disjoint events?• What is the general addition rule for union of two
events?• How do you compute P(A U B)?• What is a joint event and joint probability?• What is the general multiplication rule for any
two events?• What is meant by the conditional probability P(A|
B)?• How do you define independent events in terms
of conditional probability?
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Probability Rules from 6.2
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Venn Diagrams: Disjoint Events
AB
S
( or ) ( ) ( )P A B P A P B
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Union
P(A U B U C) = P(A) + P(B) + P(C)
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Venn Diagrams: Non-disjoint Events
A
B
S
A and B
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Venn Diagram, Non-Disjoint, in set notation
Suppose set A consists of the following
3,4,5,6
and set B consists of:
6,7,8,9
is the set 6A B
Notice that 6 is included in both sets.
So, in order to find the set (A U B) you must subtract one 6.
( ) is 3, 4,5,6,7,8,9A B
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Venn Diagrams: Non-disjoint Events
A
B
S
( or ) ( ) ( ) ( and )P A B P A P B P A B
A and B
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Joint Event
• Joint Event is the simultaneous occurrence of two events. For example, the outcome in set of A is six and the outcome in set of B is also six.
• The joint probability of a joint event is P(A and B). P(A and B) is the same as
P(A ∩ B).
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General Rule for Union of Two Events
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Example 1
• In a certain town, 40% of the people have brown hair, 25% have brown eyes, and 15% have both brown hair and brown eyes.
• A person selected at random from the town. What is the probability that the person will have brown hair or brown eyes? P( BH or BE)
• P( BH or BE) = P(BH) + P(BE) – P(BH ∩ BE) = .40 + .25 - .15 = 0.5
• What is the probability that person selected does not have brown hair or brown eyes?
• 1 – P( BH or BE) = 1 - .5 = 0.5
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Two Methods for Picturing Probabilities
• In a certain town, 40% of the people have brown hair, 25% have brown eyes, and 15% have both brown hair and brown eyes.
Venn DiagramBrown Hair Brown Eyes
.25 .10.15
.5
TableBrown Hair
Yes No Total
Brown Yes 0.15 0.25
Eyes No
Total 0.40 1.00
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Independent events
• The outcome of one trial does not influence or change the outcome of another trial.
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Multiplication Rule
• For two independent events A and B, the probability that both A and B occur is the product of the probabilities of the two events.
• P(A and B) = P(A) x P(B)
• P(A ∩ B) = P(A) x P(B)
A B
A∩B
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Independence
• Please note, that we can use the multiplication rule for independent events, to verify if two events are independent.
• If P(A and B) ≠ P(A)·P(B), then the events are not independent.
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Example 1 Continued•In a certain town, 40% of the people have brown hair, 25% have brown eyes, and 15% have both brown hair and brown eyes.
Brown Hair
Yes No Total
Brown Yes 0.15 0.25
Eyes No
Total 0.40 1.00
• What is the probability of selecting a person with brown hair? P(A) What is the probability of selecting a person with brown eyes? P(B)
• P(A) = 0.40 P(B) = 0.25
• What is the probability of selecting a person with brown hair and brown eyes? P( A and B)
• P(A and B) = 0.15
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Conditional Probability
• Lets consider the situation of finding the probability of one event under the condition that we know the results of the other event.
• The notation for conditional probability is:P(A| B)
The bar means “given the information that…”So, P(A| B) reads, “The probability of event A given event B
occurs.”
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Lets Consider Two Events That May or May Not Be Independent
If we take the equation P(A ∩ B) = P(A)P(B| A) and solve for P(B| A) we get the equation for calculating P(B| A).
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Example 1 – Independence?
• Are Events A and B independent?• P(A)●P(B) = (0.40)(0.25) = 0.10 Since P(A and B) ≠
P(A)●P(B), Events A and B are not independent.• Second Method – Test P(B| A) = P(B)
•In a certain town, 40% of the people have brown hair, 25% have brown eyes, and 15% have both brown hair and brown eyes.
Brown Hair
Yes No Total
Brown Yes 0.15 0.10 0.25
Eyes No 0.25 0.50 0.75
Total 0.40 0.60 1.00
Event A = Brown Hair
Event B = Brown Eyes
375.040.0
15.0
)(
)()|(
AP
BAPABP
Since P(A| B) ≠ 0.40, then Events are not independent.
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Example 1 – Conditional Probability
•In a certain town, 40% of the people have brown hair, 25% have brown eyes, and 15% have both brown hair and brown eyes.
Brown Hair
Yes No Total
Brown Yes 0.15 0.10 0.25
Eyes No 0.25 0.50 0.75
Total 0.40 0.60 1.00
Event A = Brown Hair
Event B = Brown Eyes
What is the P(A ∩ B) ?
Ans: P(A ∩ B) = 0.15
What is the probability that the randomly selected person has brown hair given that he has brown eyes?
Ans: P(A| B) = P(A ∩ B) / P(B) = 0.15/0.25 = 0.6
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Example 1 – Conditional Probability Continued
Brown Hair
Yes No Total
Brown Yes 0.15 0.10 0.25
Eyes No 0.25 0.50 0.75
Total 0.40 0.60 1.00
Event A = Brown Hair
Event B = Brown Eyes
What is the probability of the randomly selected person has brown eyes given the person has brown hair?
Ans: P(B| A) = P(A ∩ B) / P(A) = 0.15/0.40 = 0.375
What is the probability that the randomly selected person has neither brown hair nor brown eyes?
Ans: P( Ac ∩ Bc ) = 0.50
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Example 2
Age
18-29 30-64 65 and over Total
Married 7,842 43,808 8,270 59,920
Never Married
13,930 7,184 751 21,865
Widowed 36 2,523 8,385 10,944
Divorced 704 9,174 1,263 11,141
Total 22,512 62,689 18,669 103,870
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Example – Question 1Age
18-29 30-64 65 and over Total
Married 7,842 43,808 8,270 59,920
Never Married
13,930 7,184 751 21,865
Widowed 36 2,523 8,385 10,944
Divorced 704 9,174 1,263 11,141
Total 22,512 62,689 18,669 103,870
• A= young (between 18 and 29)• P(A)=?
– 22512/103870
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Example 2 – Question 2
Age
18-29 30-64 65 and over Total
Married 7,842 43,808 8,270 59,920
Never Married
13,930 7,184 751 21,865
Widowed 36 2,523 8,385 10,944
Divorced 704 9,174 1,263 11,141
Total 22,512 62,689 18,669 103,870
• B=married• P(B)=?
– 59920/103870
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Example 2 – Question 3Age
18-29 30-64 65 and over Total
Married 7,842 43,808 8,270 59,920
Never Married
13,930 7,184 751 21,865
Widowed 36 2,523 8,385 10,944
Divorced 704 9,174 1,263 11,141
Total 22,512 62,689 18,669 103,870
• A=is young (between 18 and 29)• B=married• P(A and B)=?
– 7842/103870
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Example 2 – Question 4Age
18-29 30-64 65 and over Total
Married 7,842 43,808 8,270 59,920
Never Married
13,930 7,184 751 21,865
Widowed 36 2,523 8,385 10,944
Divorced 704 9,174 1,263 11,141
Total 22,512 62,689 18,669 103,870
• A=is young (between 18 and 29)• B=married• P(A | B)= (Read as “the probability of A given B”)
– 7842/59920• This is known as a “conditional probability”
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Example 2 – Question 5Age
18-29 30-64 65 and over Total
Married 7,842 43,808 8,270 59,920
Never Married
13,930 7,184 751 21,865
Widowed 36 2,523 8,385 10,944
Divorced 704 9,174 1,263 11,141
Total 22,512 62,689 18,669 103,870
• A=is young (between 18 and 29)• B=married• P(B | A)= (Read as “the probability of B given A”)
– 7842/22512
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Example 2 Question 6Age
18-29 30-64 65 and over Total
Married 7,842 43,808 8,270 59,920
Never Married
13,930 7,184 751 21,865
Widowed 36 2,523 8,385 10,944
Divorced 704 9,174 1,263 11,141
Total 22,512 62,689 18,669 103,870
• P(A and B)= 7842/103870• P(A and B)= P(A)*P(B|A) 22512 7842
103870 22512
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Example 2 – Question 7Age
18-29 30-64 65 and over Total
Married 7,842 43,808 8,270 59,920
Never Married
13,930 7,184 751 21,865
Widowed 36 2,523 8,385 10,944
Divorced 704 9,174 1,263 11,141
Total 22,512 62,689 18,669 103,870
• P(A and B)= 7842/103870• P(A and B)= P(B)*P(A|B) 59920 7842
103870 59920
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Multiplication Rule Extended to Several Events
Recall the general multiplication rule for the intersection of two events.
Now, consider the intersection of three events. Remember
) B andA |P(C )A |P(B P(A) ) C B A P(
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Example 6.29 page 448• Only 5% of male high school basketball, and football players go on
to play at the college level. Of these, only 1.7% enter major league professional sports. About 40% of the athletes who compete in college and then reach the pros have a career of more than 3 years.
• Define– A = { competes in college} P(A) = 0.05– B = { competes professionally} P(B|A) = 0.017– C = { pro career longer than 3 years} P(C|A and B) = 0.40
• What is the probability that a high school athlete competes in college and then goes on to have a pro career of more than 3 years? P(A and B and C)
• P(A and B and C) = P(A)●P(B|A)●P(C|A and B) = (0.05)(0.017)(0.40) = 0.00034• SO, only 3 out of 10,000 high school athletes will have a pro-career
of more than 3 years.
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Example 3 Using Tree Diagram
• Using the same probabilities from the previous problem: – A = { competes in college} P(A) = 0.05– B = { competes professionally} P(B|A) = 0.017
• Suppose the probability of a few high school athlete enter the pro’s directly from high school (does not compete in college) is 0.0001.P(B| Ac ) = 0.0001
• What is the probability that a high school athlete will go on to the professional sport?
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Example 3Using the same probabilities from the previous problem:
A = { competes in college} P(A) = 0.05 B = { competes professionally} P(B|A) = 0.017
P(B| Ac ) = 0.0001
High School Athletes
AB
Bc
Ac
B
Bc
0.05
P(B|A) = 0.017
P(B|Ac) = 0.00010.95
0.983
0.9999
•What is the probability that a high school athlete will go on to the professional sport?• P(B) = P(A ∩ B|A) + P(Ac ∩ B|Ac) = 0.00085 + 0.000095 = 0.000945
P(A ∩ B|A) = 0.00085
P(A ∩ Bc|A) = 0.04915
P(Ac ∩ B|Ac) = 0.000095
P(Ac ∩ Bc|Ac) = 0.94991
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Tree Diagrams
• Tree diagrams combine the addition and multiplication rules.
• The multiplication rule is use to reach the end of any complete branch
• The probability of any outcome is found by adding the probabilities of all branches that are part of the event.
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A Simulation Problem
• Run a simulation for throwing two dice and finding the probability of rolling a sum of six. – Step 1: State the Problem or describe the
random event.– Step 2: State the assumptions.– Step 3: Assign digits to represent outcomes.– Step 4: Simulate many repetitions.– Step 5: State your conclusions.
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Assigning Probability to Events1 2 3 4 5 6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12
A 2 3 4 5 6 7 8 9 10 11 12
P(A) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 1/9 1/12 1/18 1/36
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Solution – Step 1: State the Problem or describe the random event.
• Find the probability of having a sum of six after a toss of two die.– Step 2: State the assumptions.
• The toss of the dice are independent.• The outcome (the sum) are not equally likely.
– Step 3: Assign digits to represent outcomes.• Use 1-36 for the 36 combinations• Number correspondence: 1 = sum of 2; 2-3 = sum of 3; 4-6 = sum of 4; 7-10 = sum of 5; 11-15 = sum of 6
– Step 4: Simulate many repetitions.• Randint(1,36,20)→L1• L1≥11 and L1≤15→L2• Sum(L2)
– Step 5: State your conclusions• The simulation resulted in approximately 30% of the rolls having a
sum of six.