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Business Statistics I

Chapter 4

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Conditional Probability

The probability of event A given thatevent B has already happened

P(A | B)

Ex: Table 4.4 (p 171) and 4.5 (p 172)

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Conditional Probability

Ex: Table 4.4 (p 171) & bottom of p 172

P(A | M) = 288 / 960 = 30%

i.e., Column Percentage

P(A | M) =

1

61 =

.24

. = .30

=( )

()

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Conditional Probability

P(M | A) = 288 / 324 = 88.889%i.e., Row Percentage

P(M | A) =

1

1

=.24

.27= .88889

=( )

()

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Conditional Probability

P(A | B) = ( )()

P(B | A) =( )

()

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

Fig 4.8 (p 173)

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Dependent Events

Dependent Events: when theprobability of an event is altered

based on knowing that some other

event occurredEx: From the officer promotion

example (pp 171-173),

P(A | M) P(A)

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Independent Events

Independent Events: when theprobability of an event is NOT

altered based on knowing that some

other event occurred

i.e.: P(A | M) = P(A)

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Multiplication Law

Addition Law is used for the union of twoevents

Multiplication Law is used for the

intersection of two events

P(A B) = P(B) * P(A | B)

= P(A) * P(B | A)

Ex: Newspapers (pp 174-175)

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Multiplication Law

Special Case for independent events:

P(A B) = P(A) * P(B)

Ex: Service Station (pp 175)

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The Game of 21

From the movie 21

http://www.youtube.com/watch?v=hcFkic2I8zU

http://www.youtube.com/watch?v=hcFkic2I8zUhttp://www.youtube.com/watch?v=hcFkic2I8zUhttp://www.youtube.com/watch?v=hcFkic2I8zUhttp://www.youtube.com/watch?v=hcFkic2I8zU

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Movie 21 Explained in WordsBreak this down into 2 options: switch, and don't switch.

If you always switch your choice, then that means you

would lose anytime you originally picked the car door (which

is 1/3 of the time), and win anytime you originally pick a

goat door (2/3 of the time)

If you don't switch, then that means you would lose anytime

you originally picked the goat door (which is 2/3 of the time),

and win anytime you originally pick the car door (1/3 of the

time).

So, you would win 2/3 of the time if you switch, and only win

1/3 of the time if you don't switch, so clearly switching is the

best play.

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Movie 21 Explained in Math

http://en.wikipedia.org/wiki/Monty_Hall_problem

Hint: Stop reading with the section Formal solution;the mathematics is beyond this class

This example from the movie introduces

Bayes Theorem

http://en.wikipedia.org/wiki/Monty_Hall_problemhttp://en.wikipedia.org/wiki/Monty_Hall_problemhttp://en.wikipedia.org/wiki/Monty_Hall_problemhttp://en.wikipedia.org/wiki/Monty_Hall_problemhttp://en.wikipedia.org/wiki/Monty_Hall_problemhttp://en.wikipedia.org/wiki/Monty_Hall_problem

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Bayes Theorem

Used whenever original probability estimates must

change because you have learned something new

Analyze based on Prior Probabilities but then new

information arrives, so we update / recalculateprobabilities based on that new information

Prior Probabilities ->New Information ->

Apply Bayes Theorem ->

Posterior Probabilities

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Bayes Theorem

Example pp 179-182

Uses Multiplication Law (p 181)

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Bayes Theorem applied to 21

Three doors; only one has a car; so P(Ai) = 1/3(i.e., the probability of any doorihaving the

car is the same as any other door)

B is the event that the host opens a door that

is not the one the contestant selects

Aj is the specific door that you choose (A1)

before you learn new information (such as

Door #3 does not have the car)

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Bayes Theorem applied to 21

To calculate the probability, we need to set

up a joint probability table for which door

the host will open compared to where the

car is, given that Ive selected Door 1

This table will allow us to calculate the

conditional probabilities

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Bayes Theorem applied to 21

Given that Ive selected Door 1, what are the joint

probabilities that the host will open door 3 given

where the car is?

Remember: the host knows where it is, but the

contestant does not

Host opens

Car is behind Door 1 Door 2 Door 3

Door 1 0

Door 2 0 0 1

Door 3 0 1 0

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Bayes Theorem applied to 21

Shown now

as a Tree

Diagram

given that

you have

selectedDoor 1:

1

2

1-18

0%

50%

50%

0%

0%

100%

0%

100%

0%

Car is behindDoor (Prob):

3

Host opensDoor:

1

2

3

1

2

3

1

2

3

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Bayes Theorem applied to 21

The denominator reads as follows:i = 1: The prob that the car is behind Door 1 times the prob

that the host opens a door other than Door 1 given than

the contestant has selected Door 1

i = 2: The prob that the car is behind Door 2 times the prob

that the host opens a door other than Door 2 given than

the contestant has selected Door 2

i = 3: The prob that the car is behind Door 3 times the prob

that the host opens a door other than Door 3 given than

the contestant has selected Door 3

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Bayes Theorem applied to 21

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Questions?