Conditional Probability

Piech, CS106A, Stanford University

Recall: S = all possible outcomes. E = the event.

Axioms of Probability

• Axiom 1: 0 £ P(E) £ 1

• Axiom 2: P(S) = 1

• Axiom 3: P(Ec) = 1 – P(E)

Aside: axiom 3 is often stated as the probability of mutually exclusive events. We’ll come back to that later in the lecture…

Recall: S = all possible outcomes. E = the event.

Axioms of Probability

• Axiom 1: 0 £ P(E) £ 1

• Axiom 2: P(S) = 1

• Axiom 3: If events E and F are mutually exclusive:

P (E [ F ) = P (E) + P (F )

Mutually Exclusive Events

P (E [ F ) = P (E) + P (F )

If events are mutually exclusive, probability of OR is simple:

7/50 4/50

P (E [ F ) = P (E) + P (F )

If events are mutually exclusive, probability of OR is simple:

7/50 4/50

P (E [ F ) =7

7/50 4/50

Mutually Exclusive Events

P (X1 [X2 [ · · · [Xn) =nX

P (Xi)

OR with Many Mutually Exclusive EventsWahoo! All my

events are mutually

exclusive

If events are mutually exclusive probability of

OR is easy!

P(Ec) = 1 – P(E)?

P(E ∪ Ec) = P(E) + P(Ec) Since E and Ec are mutually exclusive

P(S) = P(E) + P(Ec) Since everything must either be in Eor Ec

1 = P(E) + P(Ec)

P(Ec) = 1 – P(E)

Axiom 2

Rearrange

Why study probability?

• Roll two 6-sided dice, yielding values D1 and D2

• Let E be event: D1 + D2 = 4• What is P(E)?

§ |S| = 36, E = {(1, 3), (2, 2), (3, 1)}§ P(E) = 3/36 = 1/12

• Let F be event: D1 = 2• P(E, given F already observed)?

§ S = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)}§ E = {(2, 2)}§ P(E, given F already observed) = 1/6

Dice – Our Misunderstood Friends

• Two people each roll a die, yielding D1 and D2. You win if D1 + D2 = 4

• Q: What do you think is the best outcome for D1 ?

• Your Choices:§ A. 1 and 3 tie for best§ B. 1, 2 and 3 tie for best§ C. 2 is the best§ D. Other/none/more than one

• Conditional probability is probability that E occurs given that F has already occurred “Conditioning on F”

• Written as § Means “P(E, given F already observed)”§ Sample space, S, reduced to those elements

consistent with F (i.e. S Ç F)§ Event space, E, reduced to those elements

consistent with F (i.e. E Ç F)

P (E|F )

With equally likely outcomes:

P(E | F) = =

# of outcomes in E consistent with F

# of outcomes in S consistent with F

P (E|F ) =3

14⇡ 0.21

P (E) =8

50⇡ 0.16

With equally likely outcomes:

P(E | F) = =

# of outcomes in E consistent with F

# of outcomes in S consistent with F

P (E|F ) =3

14⇡ 0.21

P (E) =8

50⇡ 0.16

Shorthand notation for set intersection

(aka set “and”)

• General definition:

• Holds even when outcomes are not equally likely• Implies: P(EF) = P(E | F) P(F) (chain rule)

• What if P(F) = 0?§ P(E | F) undefined§ Congratulations! You observed the impossible!

)()()|(

FPEFPFEP =

• General definition of Chain Rule:)...( 321 nEEEEP

)...|()...|()|()( 121213121 -= nn EEEEPEEEPEEPEP

Generalized Chain Rule

Conditional Paradigm

+ Learn

Netflix and Learn

S = {Watch, Not Watch}

E = {Watch}

P(E) = ½ ?

What is the probability that a user will watch

Life is Beautiful?

Netflix and Learn

Life is Beautiful?

Netflix and Learn

P(E) = 10,234,231 / 50,923,123 = 0.20

P (E) = limn!1

n⇡ #people who watched movie

#people on Netflix

Netflix and Learn

Let E be the event that a user watched the given movie:

P(E) =0.19

P(E) =0.32

P(E) =0.20

P(E) =0.09

P(E) =0.23

* These are the actual estimates

Netflix and Learn

What is the probability that a user will watchLife is Beautiful, given they watched Amelie?

P(E|F)

P (E|F ) =P (EF )

P (F )=

#people who watched both#people on Netflix

#people who watched F#people on Netflix

P (E|F ) =P (EF )

P (F )=

Netflix and Learn

P(E|F)

P (E|F ) =P (EF )

P (F )=

Netflix and Learn

P(E|F)

P(E|F) = 0.42

P (E|F ) =P (EF )

P (F )=

#people who watched both

#people who watched F

Netflix and Learn

Let E be the event that a user watched the given movie,Let F be the event that the same user watched Amelie:

P(E|F) =0.14

P(E|F) =0.35

P(E|F) =0.20

P(E|F) =0.72

P(E|F) =0.49

Machine Learning

Machine Learning is:Probability + Data + Computers

• There are 400 students in CS109:– Probability that a random student in CS109 is a Sophomore

is 0.43– We can observe the probability that a student is both a

Sophomore and is in class– What is the conditional probability of a student coming to

class given that they are a Sophomore?• Solution:

– S is the event that a student is a sophomore– A is the event that a student is in class

Sophomores

P (A|S) = P (SA)

• Rev. Thomas Bayes (1702 –1761) was a British mathematician and Presbyterian minister

• He looked remarkably similar to Charlie Sheen§ But that’s not important right now...

Thomas Bayes

Bi-Winning!

But First!

• Say E and F are events in S

E = EF È EFc

Note: EF Ç EFc = Æ

So, P(E) = P(EF) + P(EFc)

Background Observation

Law of Total Probability

P (E) = P (EF ) + P (EFC)

= P (E|F )P (F ) + P (E|FC)P (FC)

Sample Space

Law of Total Probability

Sample Space

EB3 B4

P (E) =X

P (Bi \ E)

P (E|Bi)P (Bi)

Moment of Silence…

I want to calculate P(State of the world F | Observation E)

It seems so tricky!…

The other way around is easyP(Observation E | State of the world F)

What options to I have, chief?

P( F | E )

P( E | F )

Bayes Theorem

Bayes Theorem Want P( F | E ). Know P( E | F )

P (F |E) =P (EF )

P (E)Def. of Conditional Prob.

=P (E|F )P (F )

P (E)Chain Rule

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P (F |E) =P (EF )

=P (E|F )P (F )

P (E)Chain Rule

P (F |E) =P (EF )

=P (E|F )P (F )

P (E)Chain Rule

P (F |E) =P (EF )

=P (E|F )P (F )

P (E)Chain Rule

A little while later…

• Most common form:

• Expanded form:

Bayes Theorem

P (F |E) =P (E|F )P (F )

P (E|F )P (F ) + P (E|FC)P (FC)

P (F |E) =P (E|F )P (F )

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• A test is 98% effective at detecting H1N1§ However, test has a “false positive” rate of 1%§ 0.5% of US population has H1N1§ Let E = you test positive for H1N1 with this test§ Let F = you actually have H1N1§ What is P(F | E)?

• Solution:

P(E | F) P(F) + P(E | Fc) P(Fc)P(F | E) =

P(E | F) P(F)

(0.98)(0.005) + (0.01)(1 - 0.005)P(F | E) =

(0.98)(0.005)» 0.330

H1N1 Testing

Intuition Time

Bayes Theorem Intuition

All People

People with H1N1

All People

People who test positive

All People

People with H1N1

Conditioning on a positive result changes the sample space to this:

» 0.330

People who test positive and have H1N1

Conditioning on a positive result changes the sample space to this:

» 0.330

P(F)P(E|F)

P(F)P(E|F) + P(Fc)P(E|Fc)

People who test positive and have H1N1

All People

People with positive test

People with H1N1

Bayes Theorem IntuitionSay we have 1000 people:

5 have H1N1 and test positive, 985 do not have H1N1 and test negative.10 do not have H1N1 and test positive. » 0.333

§ Let Ec = you test negative for H1N1 with this test§ Let F = you actually have H1N1§ What is P(F | Ec)?

P(Ec | F) P(F) + P(Ec | Fc) P(Fc)P(F | Ec) = P(Ec | F) P(F)

(0.02)(0.005) + (0.99)(1 - 0.005)P(F | Ec) = (0.02)(0.005)

» 0.0001

H1N1 + H1N1 –Test + 0.98 = P(E | F) 0.01 = P(E | Fc)Test – 0.02 = P(Ec | F) 0.99 = P(Ec | Fc)

Why It’s Still Good to get Tested

Slicing Up Spam

In 2010 88% of email was spam

• Say 60% of all email is spam§ 90% of spam has a forged header§ 20% of non-spam has a forged header§ Let E = message contains a forged header§ Let F = message is spam§ What is P(F | E)?

• Solution:P(E | F) P(F) + P(E | Fc) P(Fc)

P(F | E) =P(E | F) P(F)

(0.9)(0.6) + (0.2)(0.4)P(F | E) =

(0.9)(0.6)» 0.871

Simple Spam Detection

Update Belief

Before Observation

P (L1) P (L2)

P (L5)

Update Belief

Before Observation After Observation

P (L5)

Update Belief

P (L5)

P (L5|O) =P (O|L5)P (L5)

P (L5|O)

P (L5|O) =P (O|L5)P (L5)Pi P (O|Li)P (Li)

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Update Belief

P (L5) P (L5|O)

Update Belief

P (L5) P (L5|O)

Update Belief

P (L5) P (L5|O)

Monty Hall

• Game show with 3 doors: A, B, and C

§ Behind one door is prize (equally likely to be any door)§ Behind other two doors is nothing§ We choose a door§ Then host opens 1 of other 2 doors, revealing nothing§ We are given option to change to other door

• Should we?§ Note: If we don’t switch, P(win) = 1/3 (random)

Let’s Make a Deal

• Without loss of generality, say we pick A§ P(A is winner) = 1/3

o Host opens either B or C, we always lose by switchingo P(win | A is winner, picked A, switched) = 0

§ P(B is winner) = 1/3o Host must open C (can’t open A and can’t reveal prize in B)o So, by switching, we switch to B and always wino P(win | B is winner, picked A, switched) = 1

§ P(C is winner) = 1/3o Host must open B (can’t open A and can’t reveal prize in C)o So, by switching, we switch to C and always wino P(win | C is winner, picked A, switched) = 1

§ Should always switch!o P(win | picked A, switched) = (1/3*0) + (1/3*1) + (1/3*1) = 2/3

Let’s Make a Deal

• Start with 1,000 envelopes, of which 1 is winner§ You get to choose 1 envelope

o Probability of choosing winner = 1/1000

§ Consider remaining 999 envelopeso Probability one of them is the winner = 999/1000

§ I open 998 of remaining 999 (showing they are empty)o Probability the last remaining envelope being winner =

999/1000

§ Should you switch?o Probability winning without switch =

o Probability winning with switch =

original # envelopes

original # envelopes - 1

original # envelopes

Slight Variant to Clarify

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