Chapter 3.1 Conditional Probability
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Transcript of Chapter 3.1 Conditional Probability
7/29/2019 Chapter 3.1 Conditional Probability
http://slidepdf.com/reader/full/chapter-31-conditional-probability 1/22
Mutually Exclusive Events
Events do not occur
simultaneously
A and B does not contain
any sample points, thus
P(A B) = 0
Mutually Exclusive Events
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7/29/2019 Chapter 3.1 Conditional Probability
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S
Mutually Exclusive Events
Example
Events and are Mutually Exclusive
Experiment: Draw 1 Card. Note Kind & Suit.
Outcomes
in event
Heart:
2, 3, 4, ..., A
Sample
Space:
2, 2,
2
, ..., A
Event Spade:
2, 3, 4, ..., A
2
7/29/2019 Chapter 3.1 Conditional Probability
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EXAMPLE
New England Commuter Airways recentlysupplied the following information on their
commuter flights from Boston to New York:
Arrival Frequency
Early 100
On Time 800
Late 75
Canceled 25
Total 1000
7/29/2019 Chapter 3.1 Conditional Probability
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EXAMPLE continued
If A is the event that a flight arrives early, thenP( A) = 100/1000 = 0.10
If B is the event that a flight arrives late, then:
P(B) = 75/1000 = 0.075
The probability that a flight is either early or late is:
P( A B) = P( A) + P(B) – P(A B)
= 0 .10 + 0.075 – 0 = 0.175 Note :
P(AB) = 0 because a plane cannot be early and late at the same
time so they are mutually exclusive events.
7/29/2019 Chapter 3.1 Conditional Probability
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Conditional Probability
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7/29/2019 Chapter 3.1 Conditional Probability
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Conditional Probability
1. Probability that an event will occur given that
another event has already occurred
2. If A and B are two events, then the conditional
probability of A given B is written as
3.
P(A | B)
and read as “the probabi l i ty of A given that B has
already occurred or known ”.
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7/29/2019 Chapter 3.1 Conditional Probability
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Conditional Probability
4. The formula for conditional probability is
P(A | B) = P(A and B) = P(A B)
P(B) P(B)
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7/29/2019 Chapter 3.1 Conditional Probability
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Conditional Probability Using Two –
Way Table
Draw 1 Card. Note Kind & Color . What is the
probability of getting Ace given that the card is Black?
Color
Type Red Black Total
Ace 2 2 4
Non-Ace 24 24 48
Total 26 26 52
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7/29/2019 Chapter 3.1 Conditional Probability
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Conditional Probability Using Two –
Way Table
Answer:
Color
Type Red Black Total
Ace 2 2 4
Non-Ace 24 24 48
Total 26 26 52
P(Ace Black) 2/52 2P(Ace | Black) =
P(Black) 26/52 26
= =
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7/29/2019 Chapter 3.1 Conditional Probability
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1. Event occurrence does not affect probability of another event
•
Toss 1 coin twice3. Tests for independence
• P(A | B) = P(A)
• P(A B) = P(A)*P(B)
Statistical Independence
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7/29/2019 Chapter 3.1 Conditional Probability
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Using the table then the formula, what’s the probability?
Thinking Challenge
1. P(A|D) =2. P(C|B) =
3. Are C & B
Independent?
EventEvent C D Total
A 4 2 6
B 1 3 4Total 5 5 10
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7/29/2019 Chapter 3.1 Conditional Probability
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Solution*
Using the formula, the probabilities Are:
Dependent
P(A | D) =
P(A D)
P(D)
= =
2 10
5 10
2
5
/
/
P(C | B) =P(C B)
P(B)
P(C) =5
10
= =
≠
1 10
4 10
1
4
1
4
/
/
= P(C | B)
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CHAPTER 3
Question:
A petrol station manager did a survey on his customers and found
the following information.
Payment
Customer Credit Card Cash
Regular 40 70
Non-Regular 25 45
7/29/2019 Chapter 3.1 Conditional Probability
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CHAPTER 3
Question:
(v) Assume we know that the customer is regular. What is the
probability that he will pay in credit?
(ii) Assume customer has paid in cash. What’s the probability
that he is a regular customer?
(iii) Are the two events, being a regular customer and paying
in cash, statistically independent? Explain.
7/29/2019 Chapter 3.1 Conditional Probability
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Multiplicative Rule
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Multiplicative Rule
1. Used to get compound probabilities for intersection of events
• Called joint events
2. For Dependent Events:
P(A and B) = P(A B)= P(A)*P(B|A)
= P(B)*P(A|B)
3. For Independent Events:P(A and B) = P(A B) = P(A)*P(B)
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Multiplicative Rule Example
Experiment: Draw 1 Card. Note Kind, Color & Suit.
Color
Type Red Black Total
Ace 2 2 4
Non-Ace 24 24 48
Total 26 26 52
4 2 2
52 4 52
= =
P(Ace Black) = P(Ace)∙P(Black | Ace)
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7/29/2019 Chapter 3.1 Conditional Probability
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Thinking Challenge
1. P(C B) =
2. P(B D) =
3. P(A B) =
EventEvent C D Total
A 4 2 6
B1 3 4
Total 5 5 10
Using the multiplicative rule, what’s the probability?
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7/29/2019 Chapter 3.1 Conditional Probability
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Solution*
Using the multiplicative rule, the probabilities
are:
P(C B) = P(C) P(B| C) = 5/10 * 1/5 = 1/10
P(B D) = P(B) P(D| B) = 4/10 * 3/4 = 3/10
P(A B) = P(A) P(B| A) 0 =
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7/29/2019 Chapter 3.1 Conditional Probability
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Exercise 1
Events A and B in a sample S have the following
probabilities:
P(A) = 0.4 , P(B’) = 0.3 , P(A B) = 0.2
Find
(i) P(A B).
(ii) P(A|B).(iii) Are A and B statistically independent events?
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Exercise 2
Two thousand randomly selected adults were asked if
they are in favor of or against cloning. The following
responses were given.
Opinion
Gender In Favor Against Total
Male 500 700
Female300
500
Total
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Exercise 2
If an adult is selected at random from this group, find the
probability that this adult is:
(i) in favor of cloning.
(ii) against cloning.
(iii) in favor of cloning given the person is female.
(iv) a male given that the person is against cloning.
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