dr koay probability
TRANSCRIPT
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Larson/Farber Ch. 3
pPPpPROBABILITYBy
Dr. Koay Chen Yong
Email: [email protected]
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Larson/Farber Ch. 3
Weather forecast
Games
Sports
3
Application of Probability
& Its Uses
Business
Medicine
Probability
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Larson/Farber Ch. 3
{1 2 3 4 5 6 }
{ Die is even } = { 2 4 6 }
{4}
Roll a dieProbability experiment:An action through which counts, measurements orresponses are obtained
Sample space:The set of all possible outcomes
Event:
A subset of the sample space.
Outcome:
Important Terms
The result of a single trial
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Probability Experiment: An action throughwhich counts, measurements, or responses areobtained
SampleSpace: The set of all possible outcomes
Event: A subset of the sample space.
Outcome: The result of a single trial
Choose a car from production line
Another Experiment
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Larson/Farber Ch. 3
Classical (equally probable outcomes)
Probability blood pressure will decrease
after medication
Probability the line will be busy
Empirical
Intuition
Types of Probability
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Larson/Farber Ch. 3
Two dice are rolled.
Describe thesample space.
1st roll
36 outcomes
2nd roll
Start
1 2 3 4 5 6
1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6
Tree Diagrams
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Larson/Farber Ch. 3
1,1
1,2
1,31,4
1,5
1,6
2,1
2,2
2,32,4
2,5
2,6
3,1
3,2
3,33,4
3,5
3,6
4,1
4,2
4,34,4
4,5
4,6
5,1
5,2
5,35,4
5,5
5,6
6,1
6,2
6,36,4
6,5
6,6
Find the probability the sum is 4.
Find the probability the sum is 11.
Find the probability the sum is 4 or 11.
Two dice are rolled and the sum is noted.
SampleSpace and Probabilities
P(4) =3/36 = 1/12 = 0.083
P(11) = 2/36 = 1/18= 0.056
P(4 or 11)= 5/36
= 0.139
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Larson/Farber Ch. 3
Complementary Events
The complement of event E is event E. E consistsof all the events in the sample space that are notin
event E.
The days production consists of12 cars, 5 of
which are defective. If onecar is selected at
random, find theprobability it is not defective.
E E
Solution:
P(defective) = 5/12
P(not defective) = 1 - 5/12 = 7/12 = 0.583
P(E) = 1 - P(E)
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Larson/Farber Ch. 3
Section 3.2
Independent Events
and theMultiplication Rule
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Larson/Farber Ch. 3
Two dice are rolled. Find the probability
the second die is a 4 given the first was a 4.
Original sample space: {1, 2, 3, 4, 5, 6}
Given the first die was a 4, the conditional
sample space is: {1, 2, 3, 4, 5, 6}
The conditional probability, P(B|A) = 1/6Note: For independent events,
P(A\B) = P(A) or P(B\A) = P(B)
Independent Events
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Two dice are rolled. Find the probability both are 4s.
Let A = first die is a 4 and B = second die is a 4.
P(A) = 1/6 P(B) = 1/6 and P(B\A) = 1/6
P(A and B) = 1/6 x 1/6 = 1/36 = 0.028
When two events A and B are independent, then
P (A and B) = P(A) x P(B)
Note: For independent events P(B) and P(B|A)
are the same.
Multiplication Rule
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Section 3.3
The Addition Rule
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Larson/Farber Ch. 3
Compare A and B to A or B
Thecompound event A and B means that Aand B both occur in the same trial. Use the
multiplication rule to find P(A and B).
Thecompound event A or B means either Acan occur without B, B can occur without A or
both A and B can occur. Use theaddition rule
to find P(A or B).
A B
A or BA and B
A B
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Larson/Farber Ch. 3
Mutually Exclusive Events
Two events, A and B, are mutually exclusive if
they cannot occur in the same trial.
A = a person is under 16 years old
B = a person is having driving license
A = a person was born in Kuala Lumpur
B = a person was born in Penang
A BMutually exclusive
P(A and B) = 0
When event A occurs it excludes event B in the same trial.
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Larson/Farber Ch. 3
Non-Mutually Exclusive Events
If two events can occur in the same trial, they arenon-mutually exclusive.
A = a person is under 25 years old
B = a person is a lawyer
A = a person was born in Alor Setar
B = a person watches football on TV
A B
Non-mutually exclusive
P(A and B) 0
A and B
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Larson/Farber Ch. 3
The Addition Rule
The probability that one or the other of two events willoccur is: P(A) + P(B) P(A and B)
A card is drawn from a deck. Find the
probability it is a king or it is red.
A = the card is a kingB = the card is red.
P(A) = 4/52 and P(B) = 26/52but P(A and B) = 2/52
P(A orB) = 4/52 + 26/52 2/52
= 28/52 = 0.538
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Larson/Farber Ch. 3
The Addition Rule
A card is drawn from a deck. Find the
probability the card is a king or a 10.
Let A = the card is a king B = the card is a 10.
P(A) = 4/52 and P(B) = 4/52 and P(A B) = 0
P(A orB) = 4/52 + 4/52 0 = 8/52 = 0.054
When events are mutually exclusive,
P(A or B) = P(A) + P(B)
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Larson/Farber Ch. 3
3. Probability atleast one of two events occur
P(A or B) = P(A U B)= P(A) + P(B) - P(A B)
Use Addition Rule
Addthe simple probabilities, but to prevent double counting, dont forget
to subtract the probability of both occurring.
4. Ifmutually exclusive, then P(A or B) = P(A U B)= P(A) +P(B)
as P(A B) = 0
1. For complementary events P(E') = 1 - P(E)
Subtract the probability of the event from one.
2. The probability both independentevents A and B occur
P(A and B) = P (A B) = P(A) X P(B)
UseMultiplication Rule
Summary
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Larson/Farber Ch. 3
Section 3.4
Using Tree Diagram
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If one event can occurm ways and a second event can occur
n ways, the number of ways the two events can occur insequence is m n. This rule can be extended forany number
of events occurring in a sequence.
If a meal consists of 2 choices of soup, 3 main dishes and 2 desserts,
how many different meals can be selected?
= 12 meals
Start
2
Soup
3
Main
2
Dessert
Fundamental Counting Principle
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Larson/Farber Ch. 3
Example:
ETheprobabilities that two students, A and B willpass the driving test are1/3 and 2/5 respectively.Calculate theprobability that
(a) both A and B pass the driving test,
(b) either one of them passes the driving test,
(c) at least one of them passes the driving test.B 2/5 pass Let A = A pass
pass A = A fail
1/3 3/5 fail B = B pass
A B = B fail
2/5 pass
fail
2/3 B
3/5 fail
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Larson/Farber Ch. 3
Continue:
(a) P (both A and B pass) = P(A B)
= P(A) X P(B)
= 1/3 X 2/5
= 2/15(b) P (either one of them passes)
= P(A B) + P(A B)
= (P(A) X P(B)) + (P(A) X P(B))
= (1/3 X 3/5) + (2/3 X 2/5)
= 7/15
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Larson/Farber Ch. 3
Continue:
(c) P(at least one of them passes the driving test)
= 1 P(both of them fail)
= 1 (P(A) X P(B))
= 1 ( 2/3 x 3/5)
= 1 2/5
= 3/5
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Larson/Farber Ch. 3
Group Activities
Have FUN with probabilities!