Chapter 3.1 Conditional Probability

7/29/2019 Chapter 3.1 Conditional Probability

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Mutually Exclusive Events

Events do not occur

simultaneously

A and B does not contain

any sample points, thus

P(A B) = 0


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S


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

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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



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.



Conditional Probability

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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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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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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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Conditional Probability Using Two –

Way Table

Answer:

Color


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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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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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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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



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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.



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


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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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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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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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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Chapter 3.1 Conditional Probability

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