Conditional Probabilityand Independent Events

Conditional Probability

• if we have some information about the result…use it to adjust the probability

• likelihood an event E occurs under the condition that some event F occurs

• notation: P(E | F ) "the probability of E, given F ".

• called a “conditional probability”

Given They’re Male

• If an individual is selected at random, what is the probability a sedan owner is selected, given that the owner is male?

• P( sedan owner | male ) = _______?

Smaller Sample Space

• Given the owner is male reduces the total possible outcomes to 115.

( | )

( )

n sedan maleP sedan male

n male

In general...• In terms of the probabilities, we define

( )( | )

( )

P A BP A B

P B

sedan mini-van truck totals

male .16 .10 .20 .46

female .24 .22 .08 .54 .40 .32 .28 1.00

• P( sedan owner | male ) = _______?

Compute the probability sedan mini-van truck totals

male .16 .10 .20 .46

female .24 .22 .08 .54 .40 .32 .28 1.00

( )( | ) ?

( )

P van femaleP va female

fn

P emale

( )( | ) ?

( )

P female vanP fema van

vle

P an

Compare• NOT conditional:

P( truck ) =

• Are Conditional: P( truck | male ) =

Dependent Events?

• probability of owning a truck…

• ...was affected by the knowledge the owner is male

• events "owns a truck" and "owner is male" are called dependent events.

Independent Events

• Two events E and F , are called independent if

or simply

the probability of E is unaffected by event F

Roll the Dice• Using the elements of the sample space:

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

• Compute the conditional probability: P( sum = 6 | a “4 was rolled” ) = ?

• are the events “sum = 6" and “a 4 was rolled" independent events?

“Affected”

• The events are NOT independent • the given condition does have an effect.

• That is, P(sum = 6 | 4 is rolled ) = 2/11 = 0.1818but P(sum = 6) = 5/36 = 0.1389

• These are dependent events.

Not Independent

• These are dependent events.

• As a result, P(sum = 6 and a 4 was rolled) does not equal P(sum = 6)P(a 4 was rolled) ?

20.0555

36

5 110.0424

36 36

Probability of “A and B”

• Draw two cards in succession, without replacing the first card.

• P(drawing two spades) = ________?

( ) ( | ) ( )P A B P A B P B

( )( | )

( )

P A BP A B

P B

may be written equivalently as

Multiplication Rule

P(1st ca

rd is sp

ade)

P(2nd is spade | 1st is spade)

(spade, spade)

(1 is spade 2 is spade)

(2 spade |1 spade) (1 spade)

st nd

nd st st

P

P P

Compare with “combinations approach”, ( 13C2 )( 52C2 ).

Multiplicative Law for Probability

• For two events A and B,

• But when A and B are independent events, this identity simplifies to

( ) ( | ) ( )

( | ) ( )

P A B P A B P B

P B A P A

( ) ( ) ( )P A B P A P B

Example

• In a factory, 40% of items produced come from Line 1 and others from Line 2.

• Line 1 has a defect rate of 8%. Line 2 has a defect rate of 10%.

• For randomly selected item, find probability the item is not defective. D: the selected item is defective (i.e., ~D means not defective)

The Decision Tree

Line 1

Line 2

defective

defective

not defective

not defective

1 1 2 2(~ ) (~ | ) ( ) (~ | ) ( )P D P D L P L P D L P L

(~ ) (0.92)(0.40) (0.90)(0.60) 0.908P D

The Two Lines

• ~D: the selected item is not defective.

S

L1 L2

~D

1 2(~ ) (~ ) (~ )P D P D L P D L

1 1 2 2(~ | ) ( ) (~ | ) ( )P D L P L P D L P L

Total Probability

1 2

1 2

1 2

For the partition { , , , } of the sample space ,

we may write ( ) ( ) ( )

and so

( ) ( ) ( ) ( ).

k

k

k

B B B S

A A B A B A B

P A P A B P A B P A B

S

B1 B2 Bk…

A

1 1 2 2

or equivalently,

( ) ( | ) ( ) ( | ) ( ) ( | ) ( ).k kP A P A B P B P A B P B P A B P B

Total Probability

B1

B2

B3

A

A

A

A

A

A

P(A|B1)P(B1)

P(A|B2)P(B2)

P(A|B3)P(B3)

1 1 2 2 3 3

( )

( | ) ( ) ( | ) ( ) ( | ) ( ).

P A

P A B P B P A B P B P A B P B

Bayes’ Theorem follows…

( ) ( | ) ( )( | )

( ) ( )j j j

j

P A B P A B P BP B A

P A P A

1 1Since ( ) ( | ) ( ) ( | ) ( ),

we also havek kP A P A B P B P A B P B

1 1

( | ) ( )

( | ) ( ) ( | ) ( )j j

k k

P A B P B

P A B P B P A B P B

Bayes’

B1

B2

B3

A

A

A

A

A

A

P(A|B1)P(B1)

P(A|B2)P(B2)

P(A|B3)P(B3)

2 22

1 1 2 2 3 3

( | ) ( )( | )

( | ) ( ) ( | ) ( ) ( | ) ( )

P A B P BP B A

P A B P B P A B P B P A B P B

Back at the factory…• For randomly selected item, find probability

it came from Line 1, given the item is not defective. P( L1 | W ) =

Line 1

Line 2

defective

defective

not defective

not defective

The 3 Urns

• Three urns contain colored balls. Urn Red White Blue 1 3 4 1 2 1 2 3 3 4 3 2

• An urn is selected at random and one ball is randomly selected from the urn.

• Given that the ball is red, what is the probability it came from urn #2 ?