Conditional Probability and Independence

Conditional Probability and Independence

Christopher Croke

University of Pennsylvania

Math 115

Christopher Croke Calculus 115

Probability

The probability of an event depends on the sample space.

Problem: In a group of 30 athletes 12 are women, 18 areswimmers, and 10 are neither. A person is chosen at random.What is the probability that it is a female swimmer?Here we find (from inclusion/exclusion) that #(S ∩W ) = 10 so

Pr(S ∩W ) =#(S ∩W )

total#=

10

30=

1

3.

Suppose we know we have chosen a woman. What is theprobability that she is a swimmer? Probability is

#(S ∩W )

#Sample Space=

#(S ∩W )

#W=

10

12=

5

6.


Probability

The probability of an event depends on the sample space.Problem: In a group of 30 athletes 12 are women, 18 areswimmers, and 10 are neither. A person is chosen at random.

What is the probability that it is a female swimmer?Here we find (from inclusion/exclusion) that #(S ∩W ) = 10 so

Pr(S ∩W ) =#(S ∩W )

total#=

10

30=

1

3.


#(S ∩W )

#Sample Space=

#(S ∩W )

#W=

10

12=

5

6.


Probability

The probability of an event depends on the sample space.Problem: In a group of 30 athletes 12 are women, 18 areswimmers, and 10 are neither. A person is chosen at random.What is the probability that it is a female swimmer?

Here we find (from inclusion/exclusion) that #(S ∩W ) = 10 so

Pr(S ∩W ) =#(S ∩W )

total#=

10

30=

1

3.


#(S ∩W )

#Sample Space=

#(S ∩W )

#W=

10

12=

5

6.


Probability

The probability of an event depends on the sample space.Problem: In a group of 30 athletes 12 are women, 18 areswimmers, and 10 are neither. A person is chosen at random.What is the probability that it is a female swimmer?Here we find (from inclusion/exclusion) that #(S ∩W ) = 10 so

Pr(S ∩W ) =#(S ∩W )

total#=

10

30=

1

3.


#(S ∩W )

#Sample Space=

#(S ∩W )

#W=

10

12=

5

6.


Probability


Pr(S ∩W ) =#(S ∩W )

total#=

10

30=

1

3.

Suppose we know we have chosen a woman. What is theprobability that she is a swimmer?

Probability is

#(S ∩W )

#Sample Space=

#(S ∩W )

#W=

10

12=

5

6.


Probability


Pr(S ∩W ) =#(S ∩W )

total#=

10

30=

1

3.


#(S ∩W )

#Sample Space=

#(S ∩W )

#W=

10

12=

5

6.


Probability


Pr(S ∩W ) =#(S ∩W )

total#=

10

30=

1

3.


#(S ∩W )

#Sample Space=

#(S ∩W )

#W=

10

12=

5

6.


Conditional Probability

The conditional probability of E given F denoted Pr(E |F )

isgiven by

Pr(E |F ) =Pr(E ∩ F )

Pr(F ).

In the previous example (where there were equally likely outcomes)we saw:

Pr(E |F ) =#(E ∩ F )

#F

But this is equal to

#(E ∩ F )

(total#)

(total#)

#F=

Pr(E ∩ F )

Pr(F ).



The conditional probability of E given F denoted Pr(E |F ) isgiven by

Pr(E |F ) =Pr(E ∩ F )

Pr(F ).


Pr(E |F ) =#(E ∩ F )

#F


#(E ∩ F )

(total#)

(total#)

#F=

Pr(E ∩ F )

Pr(F ).




Pr(E |F ) =Pr(E ∩ F )

Pr(F ).


Pr(E |F ) =#(E ∩ F )

#F


#(E ∩ F )

(total#)

(total#)

#F=

Pr(E ∩ F )

Pr(F ).




Pr(E |F ) =Pr(E ∩ F )

Pr(F ).


Pr(E |F ) =#(E ∩ F )

#F


#(E ∩ F )

(total#)

(total#)

#F=

Pr(E ∩ F )

Pr(F ).


Why does this make sense?

Consider a large number N of trials. We expect the number ofoutcomes that shows up in F to be about NPr(F ). Of these weexpect about NPr(E ∩ F ) also to land in E . So the fraction ofthose that landed in F that also landed in E is

NPr(E ∩ F )

NPr(F )=

Pr(E ∩ F )

Pr(F )= Pr(E |F ).

PRODUCT RULE: Pr(E ∩ F ) = Pr(F ) · Pr(E |F ).In our example we see 1

3 = 1230 ·

56 .


Why does this make sense?Consider a large number N of trials. We expect the number ofoutcomes that shows up in F to be about NPr(F ).

Of these weexpect about NPr(E ∩ F ) also to land in E . So the fraction ofthose that landed in F that also landed in E is

NPr(E ∩ F )

NPr(F )=

Pr(E ∩ F )

Pr(F )= Pr(E |F ).


3 = 1230 ·

56 .


Why does this make sense?Consider a large number N of trials. We expect the number ofoutcomes that shows up in F to be about NPr(F ). Of these weexpect about NPr(E ∩ F ) also to land in E .

So the fraction ofthose that landed in F that also landed in E is

NPr(E ∩ F )

NPr(F )=

Pr(E ∩ F )

Pr(F )= Pr(E |F ).


3 = 1230 ·

56 .


Why does this make sense?Consider a large number N of trials. We expect the number ofoutcomes that shows up in F to be about NPr(F ). Of these weexpect about NPr(E ∩ F ) also to land in E . So the fraction ofthose that landed in F that also landed in E is

NPr(E ∩ F )

NPr(F )=

Pr(E ∩ F )

Pr(F )= Pr(E |F ).


3 = 1230 ·

56 .



NPr(E ∩ F )

NPr(F )=

Pr(E ∩ F )

Pr(F )= Pr(E |F ).

PRODUCT RULE: Pr(E ∩ F ) = Pr(F ) · Pr(E |F ).

In our example we see 13 = 12

30 ·56 .



NPr(E ∩ F )

NPr(F )=

Pr(E ∩ F )

Pr(F )= Pr(E |F ).


3 = 1230 ·

56 .


Problem: Two students are chosen, one after the other, from agroup of 50 students, 20 of who are freshmen and 30 of who aresophomores.

a) What is the probability that the first is a freshman and thesecond is a sophomore?b) If three are chosen, what is the probability that the first is asophomore, and the next two are freshmen?we will need the Generalized Product Rule:

Pr(E1 ∩ E2 ∩ E3) = Pr(E1) · Pr(E2|E1) · Pr(E3|E1 ∩ E2).


Problem: Two students are chosen, one after the other, from agroup of 50 students, 20 of who are freshmen and 30 of who aresophomores.a) What is the probability that the first is a freshman and thesecond is a sophomore?

b) If three are chosen, what is the probability that the first is asophomore, and the next two are freshmen?we will need the Generalized Product Rule:



Problem: Two students are chosen, one after the other, from agroup of 50 students, 20 of who are freshmen and 30 of who aresophomores.a) What is the probability that the first is a freshman and thesecond is a sophomore?b) If three are chosen, what is the probability that the first is asophomore, and the next two are freshmen?

we will need the Generalized Product Rule:



Problem: Two students are chosen, one after the other, from agroup of 50 students, 20 of who are freshmen and 30 of who aresophomores.a) What is the probability that the first is a freshman and thesecond is a sophomore?b) If three are chosen, what is the probability that the first is asophomore, and the next two are freshmen?we will need the Generalized Product Rule:



Independence

Two events E and F are said to be independent ifPr(E ) = Pr(E |F ) (as long as Pr(F ) 6= 0).

This is the same as (the official definition):

Pr(E ∩ F ) = Pr(E ) · Pr(F ).

Note this also means Pr(F ) = Pr(F |E ).Example: Roll a die two times.Let E be “got a 1 on first roll”.Let F be “got a 3 on second roll”.Check that these are independent.


Independence

Two events E and F are said to be independent ifPr(E ) = Pr(E |F ) (as long as Pr(F ) 6= 0).This is the same as (the official definition):

Pr(E ∩ F ) = Pr(E ) · Pr(F ).



Independence


Pr(E ∩ F ) = Pr(E ) · Pr(F ).

Note this also means Pr(F ) = Pr(F |E ).

Example: Roll a die two times.Let E be “got a 1 on first roll”.Let F be “got a 3 on second roll”.Check that these are independent.


Independence


Pr(E ∩ F ) = Pr(E ) · Pr(F ).



Problem: A card is to be drawn from a full deck. Consider theevents:E=“the card is a 4”.F=“the card is a spade”a) Are these independent events?

b) Answer the same question when the original deck was missingthe 7 of clubs.c) Answer the same question when the original deck was missingthe ace of spades and all the clubs and the ace and king ofdiamonds.

If E and F are independent then so are E c and F c . Also E and F c

are independent, etc.

Pr(E∩F c) = Pr(E−E∩F ) = Pr(E )−Pr(E∩F ) = Pr(E )−Pr(E )Pr(F ) =

Pr(E )(1− Pr(F ))) = Pr(E )Pr(F c)


Problem: A card is to be drawn from a full deck. Consider theevents:E=“the card is a 4”.F=“the card is a spade”a) Are these independent events?b) Answer the same question when the original deck was missingthe 7 of clubs.

c) Answer the same question when the original deck was missingthe ace of spades and all the clubs and the ace and king ofdiamonds.




Pr(E )(1− Pr(F ))) = Pr(E )Pr(F c)


Problem: A card is to be drawn from a full deck. Consider theevents:E=“the card is a 4”.F=“the card is a spade”a) Are these independent events?b) Answer the same question when the original deck was missingthe 7 of clubs.c) Answer the same question when the original deck was missingthe ace of spades and all the clubs and the ace and king ofdiamonds.




Pr(E )(1− Pr(F ))) = Pr(E )Pr(F c)



If E and F are independent then so are E c and F c .

Also E and F c



Pr(E )(1− Pr(F ))) = Pr(E )Pr(F c)






Pr(E )(1− Pr(F ))) = Pr(E )Pr(F c)





Pr(E∩F c) = Pr(E−E∩F )

= Pr(E )−Pr(E∩F ) = Pr(E )−Pr(E )Pr(F ) =

Pr(E )(1− Pr(F ))) = Pr(E )Pr(F c)





Pr(E∩F c) = Pr(E−E∩F ) = Pr(E )−Pr(E∩F )

= Pr(E )−Pr(E )Pr(F ) =

Pr(E )(1− Pr(F ))) = Pr(E )Pr(F c)





Pr(E∩F c) = Pr(E−E∩F ) = Pr(E )−Pr(E∩F ) = Pr(E )−Pr(E )Pr(F )

=

Pr(E )(1− Pr(F ))) = Pr(E )Pr(F c)






Pr(E )(1− Pr(F )))

= Pr(E )Pr(F c)






Pr(E )(1− Pr(F ))) = Pr(E )Pr(F c)


Independence of a collection of events

A collection A1,A2,A3, ...,An of events are independent if for anysubcollection Ai1 ,Ai2 ,Ai3 , ...,Aik we have

Pr(Ai1 ∩ Ai2 ∩ Ai3 ∩ ... ∩ Aik ) = Pr(Ai1)Pr(Ai2)Pr(Ai3)...Pr(Aik ).

For example if our collection has 3 events E , F , and G then weneed:

Pr(E ∩ F ) = Pr(E )Pr(F )

Pr(E ∩ G ) = Pr(E )Pr(G )

Pr(F ∩ G ) = Pr(F )Pr(G )

Pr(E ∩ F ∩ G ) = Pr(E )Pr(F )Pr(G )

It is not enough that the events are pairwise independent.
































Problem: If E, F and G are three independent events withPr(E ) = .5, Pr(F ) = .4, and Pr(G ) = .3 calculate:a) Pr(E ∩ F ∩ G ).b) Pr(E ∩ G c).c) Pr(E ∩ (F ∪ G )c).

d) Pr(E ∪ (F ∪ G )c).

Problem:(Inspecting) A machine produces defective items with aprobability p.a) If 10 items are chosen at random what is the probability thatexactly 3 are defective?b) What is the probability of finding at least one defective item inthe 10 chosen.c) If we observe the items one at a time as they come off the line,what is the probability that the 3rd defective item is the 10th itemobserved?


Problem: If E, F and G are three independent events withPr(E ) = .5, Pr(F ) = .4, and Pr(G ) = .3 calculate:a) Pr(E ∩ F ∩ G ).b) Pr(E ∩ G c).c) Pr(E ∩ (F ∪ G )c).d) Pr(E ∪ (F ∪ G )c).




Problem:(Inspecting) A machine produces defective items with aprobability p.a) If 10 items are chosen at random what is the probability thatexactly 3 are defective?

b) What is the probability of finding at least one defective item inthe 10 chosen.c) If we observe the items one at a time as they come off the line,what is the probability that the 3rd defective item is the 10th itemobserved?



Problem:(Inspecting) A machine produces defective items with aprobability p.a) If 10 items are chosen at random what is the probability thatexactly 3 are defective?b) What is the probability of finding at least one defective item inthe 10 chosen.

c) If we observe the items one at a time as they come off the line,what is the probability that the 3rd defective item is the 10th itemobserved?





Using tree diagrams

You can use tree diagrams when performing a sequence ofexperiments.

So the probabilities on the edges are conditional probabilities.From the diagram we see

Pr(O2 ∩ Oa) = 0.3 · 0.5 = 0.15.


Using tree diagrams



Pr(O2 ∩ Oa) = 0.3 · 0.5 = 0.15.


Using tree diagrams


So the probabilities on the edges are conditional probabilities.

From the diagram we see

Pr(O2 ∩ Oa) = 0.3 · 0.5 = 0.15.


Using tree diagrams



Pr(O2 ∩ Oa) = 0.3 · 0.5 = 0.15.


Problem:

A city of 100,000 people is broken into 4 police precincts of unequalsize (call them Pre1, Pre2, Pre3, Pre4). The population of Pr1 is10,000, of Pr2 is 20,000, of Pr3 is 30,000, and of Pr4 is 40,000. Areview of crime recording shows that mistakes have been made.

20% of records in Pr1 contain errors.5% of records in Pr2 contain errors.10% of records in Pr3 contain errors.5% of records in Pr4 contain errors.a) Draw a tree diagram to describe these results.b) What is the probability that a record has an error and is in Pre3?c) What is the probability that a record chosen at random has anerror?d) What is the probability that a record is from Pre3 given that itcontains an error?


Problem:

A city of 100,000 people is broken into 4 police precincts of unequalsize (call them Pre1, Pre2, Pre3, Pre4). The population of Pr1 is10,000, of Pr2 is 20,000, of Pr3 is 30,000, and of Pr4 is 40,000. Areview of crime recording shows that mistakes have been made.20% of records in Pr1 contain errors.5% of records in Pr2 contain errors.10% of records in Pr3 contain errors.5% of records in Pr4 contain errors.a) Draw a tree diagram to describe these results.

b) What is the probability that a record has an error and is in Pre3?c) What is the probability that a record chosen at random has anerror?d) What is the probability that a record is from Pre3 given that itcontains an error?


Problem:

A city of 100,000 people is broken into 4 police precincts of unequalsize (call them Pre1, Pre2, Pre3, Pre4). The population of Pr1 is10,000, of Pr2 is 20,000, of Pr3 is 30,000, and of Pr4 is 40,000. Areview of crime recording shows that mistakes have been made.20% of records in Pr1 contain errors.5% of records in Pr2 contain errors.10% of records in Pr3 contain errors.5% of records in Pr4 contain errors.a) Draw a tree diagram to describe these results.b) What is the probability that a record has an error and is in Pre3?

c) What is the probability that a record chosen at random has anerror?d) What is the probability that a record is from Pre3 given that itcontains an error?


Problem:

A city of 100,000 people is broken into 4 police precincts of unequalsize (call them Pre1, Pre2, Pre3, Pre4). The population of Pr1 is10,000, of Pr2 is 20,000, of Pr3 is 30,000, and of Pr4 is 40,000. Areview of crime recording shows that mistakes have been made.20% of records in Pr1 contain errors.5% of records in Pr2 contain errors.10% of records in Pr3 contain errors.5% of records in Pr4 contain errors.a) Draw a tree diagram to describe these results.b) What is the probability that a record has an error and is in Pre3?c) What is the probability that a record chosen at random has anerror?

d) What is the probability that a record is from Pre3 given that itcontains an error?


Problem:

A city of 100,000 people is broken into 4 police precincts of unequalsize (call them Pre1, Pre2, Pre3, Pre4). The population of Pr1 is10,000, of Pr2 is 20,000, of Pr3 is 30,000, and of Pr4 is 40,000. Areview of crime recording shows that mistakes have been made.20% of records in Pr1 contain errors.5% of records in Pr2 contain errors.10% of records in Pr3 contain errors.5% of records in Pr4 contain errors.a) Draw a tree diagram to describe these results.b) What is the probability that a record has an error and is in Pre3?c) What is the probability that a record chosen at random has anerror?d) What is the probability that a record is from Pre3 given that itcontains an error?


Problem:

A city of 100,000 people is broken into 4 police precincts of unequalsize (call them Pre1, Pre2, Pre3, Pre4). The population of Pr1 is10,000, of Pr2 is 20,000, of Pr3 is 30,000, and of Pr4 is 40,000. Areview of crime recording shows that mistakes have been made.20% of records in Pr1 contain errors.5% of records in Pr2 contain errors.10% of records in Pr3 contain errors.5% of records in Pr4 contain errors.a) Draw a tree diagram to describe these results.b) What is the probability that a record has an error and is in Pre3?c) What is the probability that a record chosen at random has anerror?d) What is the probability that a record is from Pre3 given that itcontains an error?


Problem: There is a (very good) test for TB that will test positivefor TB 98% of the time if a person has TB while it will only testpositive 1% of the time if the person doesn’t have TB. Given thatonly 0.02% of the population has TB, what is the probability that apatient has TB if she tests positive (i.e. a so called false positive)?

Trees are not always symmetric!

Problem: A crate of apples contains 3 bad apples and 7 goodapples. Apples are chosen until a good one is picked. What is theprobability that it takes at least 3 picks to get a good apple?










Bayes Theorem

Return to the Precincts problem with the files that had errors. Theevents Pre1, Pre2, Pre3 and Pre4 were mutually exclusive eventswhose union Pre1 ∪ Pre2 ∪ Pre3 ∪ Pre4 = S was the whole samplespace (i.e. they form a partition of S).

We saw that Pr(E ∩ Pre3) = Pr(Pre3)Pr(E |Pre3).While Pr(E ) = Pr(Pre1)Pr(E |Pre1) + Pr(Pre2)Pr(E |Pre2) +Pr(Pre3)Pr(E |Pre3) + Pr(Pre4)Pr(E |Pre4).

Hence Pr(Pre3|E ) = Pr(E∩Pre3)Pr(E) =

Pr(Pre3)Pr(E |Pre3)Pr(Pre1)Pr(E |Pre1)+Pr(Pre2)Pr(E |Pre2)+Pr(Pre3)Pr(E |Pre3)+Pr(Pre4)Pr(E |Pre4) .

This works in general:


Bayes Theorem

Return to the Precincts problem with the files that had errors. Theevents Pre1, Pre2, Pre3 and Pre4 were mutually exclusive eventswhose union Pre1 ∪ Pre2 ∪ Pre3 ∪ Pre4 = S was the whole samplespace (i.e. they form a partition of S).We saw that Pr(E ∩ Pre3) = Pr(Pre3)Pr(E |Pre3).

While Pr(E ) = Pr(Pre1)Pr(E |Pre1) + Pr(Pre2)Pr(E |Pre2) +Pr(Pre3)Pr(E |Pre3) + Pr(Pre4)Pr(E |Pre4).





Bayes Theorem

Return to the Precincts problem with the files that had errors. Theevents Pre1, Pre2, Pre3 and Pre4 were mutually exclusive eventswhose union Pre1 ∪ Pre2 ∪ Pre3 ∪ Pre4 = S was the whole samplespace (i.e. they form a partition of S).We saw that Pr(E ∩ Pre3) = Pr(Pre3)Pr(E |Pre3).While Pr(E ) = Pr(Pre1)Pr(E |Pre1) + Pr(Pre2)Pr(E |Pre2) +Pr(Pre3)Pr(E |Pre3) + Pr(Pre4)Pr(E |Pre4).





Bayes Theorem






Bayes Theorem






Bayes’ Theorem

BAYES’ THEOREM If B1, B2,....,Bn are mutually exclusiveevents whose union is the whole sample space then for any event Awe have Pr(Bi |A) is

Pr(Bi )Pr(A|Bi )

Pr(B1)Pr(A|B1) + Pr(B2)Pr(A|B2) + ... + Pr(Bn)Pr(A|Bn).

Problem: A refrigerator manufacturer has plants in five cities.The following chart describes the daily production and therejection rate in the five cities.City Units output failure rateAtlanta 50 .03Boston 100 .02Chicago 400 .02Detroit 400 .03Eugene 50 .01What is the probability that a refrigerator was manufactured inChicago given that it was rejected?


Bayes’ Theorem


Pr(Bi )Pr(A|Bi )




Bayes’ Theorem


Pr(Bi )Pr(A|Bi )


Problem: A refrigerator manufacturer has plants in five cities.The following chart describes the daily production and therejection rate in the five cities.City Units output failure rateAtlanta 50 .03Boston 100 .02Chicago 400 .02Detroit 400 .03Eugene 50 .01

What is the probability that a refrigerator was manufactured inChicago given that it was rejected?


Bayes’ Theorem


Pr(Bi )Pr(A|Bi )




Bayes’ Theorem

Problem: It is observed that at any intersection 80% of the carsthat turn use their turn signals when turning. At a certainintersection 85% of cars make a turn. If at this intersection you arebehind a car not using a turn signal what is the probability that itwill turn anyway? (We assume that cars going straight never use asignal).


Prior and Posterior Probabilities

Consider Pr(B) and Pr(B|A).

Pr(B) is a Prior Probability (Because it is the probability of Bwith no other information.)Pr(B|A) ia a Posterior Probability (Because it is the probabilityof B *after* we know A holds.)Example: We have two coins. Coin 1 is a fair coin while Coin 2has two heads. We will select a coin randomly and toss it.Let B1 be the event that the coin is fair and B2 be the event thatthe coin is two headed.The prior probabilities are Pr(B1) = Pr(B2) = 1

2 .

Now flip the coin. Say a head comes up (event H1). What are theposterior probabilities Pr(B1|H1) and Pr(B2|H1)? Flip the coinagain and say a head comes up again (event H2). What are theposterior probabilities?



Consider Pr(B) and Pr(B|A).Pr(B) is a Prior Probability (Because it is the probability of Bwith no other information.)

Pr(B|A) ia a Posterior Probability (Because it is the probabilityof B *after* we know A holds.)Example: We have two coins. Coin 1 is a fair coin while Coin 2has two heads. We will select a coin randomly and toss it.Let B1 be the event that the coin is fair and B2 be the event thatthe coin is two headed.The prior probabilities are Pr(B1) = Pr(B2) = 1

2 .




Consider Pr(B) and Pr(B|A).Pr(B) is a Prior Probability (Because it is the probability of Bwith no other information.)Pr(B|A) ia a Posterior Probability (Because it is the probabilityof B *after* we know A holds.)

Example: We have two coins. Coin 1 is a fair coin while Coin 2has two heads. We will select a coin randomly and toss it.Let B1 be the event that the coin is fair and B2 be the event thatthe coin is two headed.The prior probabilities are Pr(B1) = Pr(B2) = 1

2 .




Consider Pr(B) and Pr(B|A).Pr(B) is a Prior Probability (Because it is the probability of Bwith no other information.)Pr(B|A) ia a Posterior Probability (Because it is the probabilityof B *after* we know A holds.)Example: We have two coins. Coin 1 is a fair coin while Coin 2has two heads. We will select a coin randomly and toss it.

Let B1 be the event that the coin is fair and B2 be the event thatthe coin is two headed.The prior probabilities are Pr(B1) = Pr(B2) = 1

2 .




Consider Pr(B) and Pr(B|A).Pr(B) is a Prior Probability (Because it is the probability of Bwith no other information.)Pr(B|A) ia a Posterior Probability (Because it is the probabilityof B *after* we know A holds.)Example: We have two coins. Coin 1 is a fair coin while Coin 2has two heads. We will select a coin randomly and toss it.Let B1 be the event that the coin is fair and B2 be the event thatthe coin is two headed.

The prior probabilities are Pr(B1) = Pr(B2) = 12 .




Consider Pr(B) and Pr(B|A).Pr(B) is a Prior Probability (Because it is the probability of Bwith no other information.)Pr(B|A) ia a Posterior Probability (Because it is the probabilityof B *after* we know A holds.)Example: We have two coins. Coin 1 is a fair coin while Coin 2has two heads. We will select a coin randomly and toss it.Let B1 be the event that the coin is fair and B2 be the event thatthe coin is two headed.The prior probabilities are Pr(B1) = Pr(B2) = 1

2 .





2 .

Now flip the coin. Say a head comes up (event H1).

What are theposterior probabilities Pr(B1|H1) and Pr(B2|H1)? Flip the coinagain and say a head comes up again (event H2). What are theposterior probabilities?




2 .

Now flip the coin. Say a head comes up (event H1). What are theposterior probabilities Pr(B1|H1) and Pr(B2|H1)?

Flip the coinagain and say a head comes up again (event H2). What are theposterior probabilities?




2 .



Conditional version of Bayes’ Theorem

There are two ways to solve.

The first is like before:

Pr(B1|H1∩H2) =Pr(B1)Pr(H1 ∩ H2|B1)

Pr(B1)Pr(H1 ∩ H2|B1) + Pr(B2)Pr(H1 ∩ H2|B2)=

=12 ·

14

12 ·

14 + 1

2 · 1=

1

5.

The other way is to use a conditional version of Bayes’ Theorem.:

Pr(Bi |A ∩ C ) =Pr(Bi |C )Pr(A|Bi ∩ C )

Σnj=1Pr(Bj |C )Pr(A|Bj ∩ C )

.

in our case (C = H1 and A = H2) we get:

13 ·

12

13 ·

12 + 2

3 · 1=

1

5.



There are two ways to solve. The first is like before:

Pr(B1|H1∩H2) =Pr(B1)Pr(H1 ∩ H2|B1)


=12 ·

14

12 ·

14 + 1

2 · 1=

1

5.




.


13 ·

12

13 ·

12 + 2

3 · 1=

1

5.




Pr(B1|H1∩H2) =Pr(B1)Pr(H1 ∩ H2|B1)


=12 ·

14

12 ·

14 + 1

2 · 1

=1

5.




.


13 ·

12

13 ·

12 + 2

3 · 1=

1

5.




Pr(B1|H1∩H2) =Pr(B1)Pr(H1 ∩ H2|B1)


=12 ·

14

12 ·

14 + 1

2 · 1=

1

5.




.


13 ·

12

13 ·

12 + 2

3 · 1=

1

5.




Pr(B1|H1∩H2) =Pr(B1)Pr(H1 ∩ H2|B1)


=12 ·

14

12 ·

14 + 1

2 · 1=

1

5.




.


13 ·

12

13 ·

12 + 2

3 · 1=

1

5.




Pr(B1|H1∩H2) =Pr(B1)Pr(H1 ∩ H2|B1)


=12 ·

14

12 ·

14 + 1

2 · 1=

1

5.




.


13 ·

12

13 ·

12 + 2

3 · 1=

1

5.


For two events A,B they are independent if Pr(A) = Pr(A|B).

We can see {A1,A2, ...,An} independent if for any two disjointsubsets {i1, i2, ..., ik} and {j1, j2, ..., jl} of {1, 2, ..., n} we have:

Pr(Ai1 ∩Ai2 ∩ ...∩Aik ) = Pr(Ai1 ∩Ai2 ∩ ...∩Aik |Aj1 ∩Aj2 ∩ ...∩Ajl ).

For example:

Pr(A2 ∩ A6 ∩ A8) = Pr(A2 ∩ A6 ∩ A8|A1 ∩ A3 ∩ A5 ∩ A7).

{A1,A2, ...,An} are conditionally independent given B if forevery subset {Ai1 ,Ai2 , ...,Aik} we have

Pr(Ai1 ∩ Ai2 ∩ ... ∩ Aik |B) = Pr(Ai1 |B)Pr(Ai2 |B)...Pr(Aik |B).


For two events A,B they are independent if Pr(A) = Pr(A|B).We can see {A1,A2, ...,An} independent if for any two disjointsubsets {i1, i2, ..., ik} and {j1, j2, ..., jl} of {1, 2, ..., n} we have:


For example:

Pr(A2 ∩ A6 ∩ A8) = Pr(A2 ∩ A6 ∩ A8|A1 ∩ A3 ∩ A5 ∩ A7).




For two events A,B they are independent if Pr(A) = Pr(A|B).We can see {A1,A2, ...,An} independent if for any two disjointsubsets {i1, i2, ..., ik} and {j1, j2, ..., jl} of {1, 2, ..., n} we have:


For example:

Pr(A2 ∩ A6 ∩ A8) = Pr(A2 ∩ A6 ∩ A8|A1 ∩ A3 ∩ A5 ∩ A7).




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Conditional Probability and Independence