4. Conditional Probability P( ) CSE 312 Winter 2011 W.L. Ruzzo

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1 4. Conditional Probability P( ) CSE 312 Winter 2011 W.L. Ruzzo

2 conditional probability Conditional probability of E given F: probability that E occurs given that F has already occurred. Conditioning on F Written as P(E F) Means P(E, given F already observed) Sample space S reduced to those elements consistent with F (i.e. S F) Event space E reduced to those elements consistent with F (i.e. E F) With equally likely outcomes,

3 conditional probability General defn: where P(F) > 0 Holds even when outcomes are not equally likely. What if P(F) = 0? P(E F) undefined: (you can t observe the impossible) Implies: P(EF) = P(E F) P(F) (chain rule) General definition of Chain Rule:

4 coin flipping Suppose you flip two coins & all outcomes are equally likely. What is the probability that both flips land on heads if The first flip lands on heads? Let B = {HH} and F = {HH, HT} P(B F) = P(BF)/P(F) = P({HH})/P({HH, HT}) = (1/4)/(2/4) = 1/2 At least one of the two flips lands on heads? Let A = {HH, HT, TH} P(B A) = P(BA)/P(A) = P({HH})/P({HH, HT, TH}) = (1/4)/(3/4) = 1/3

5 sending bit strings

6 sending bit strings Bit string with m 0 s and n 1 s sent on the network All distinct arrangements of bits equally likely E = first bit received is a 1 F = k of first r bits received are 1 s What s P(E F)? Solution 1:

7 sending bit strings Bit string with m 0 s and n 1 s sent on the network All distinct arrangements of bits equally likely E = first bit received is a 1 F = k of first r bits received are 1 s Solution 2: Observe: P(E F) = P(picking one of k 1 s out of r bits) So: P(E F) = k/r

8 piling cards

9 piling cards Deck of 52 cards randomly divided into 4 piles 13 cards per pile Compute P(each pile contains an ace) Solution: E 1 = { in any one pile } E 2 = { and in different piles } E 3 = { different piles } E 4 = { all four aces in different piles } Compute P(E 1 E 2 E 3 E 4 )

10 piling cards E 1 = { in any one pile } E 2 = { and in different piles } E 3 = { different piles } E 4 = { all four aces in different piles } P(E 1 E 2 E 3 E 4 ) = P(E 1 ) P(E 2 E 1 ) P(E 3 E 1 E 2 ) P(E 4 E 1 E 2 E 3 )

11 piling cards E 1 = { in any one pile } E 2 = { and in different piles } E 3 = { different piles } E 4 = { all four aces in different piles } P(E 1 ) = 1 P(E 2 E 1 ) = 39/51 (39 cards not in AH pile) P(E 3 E 1 E 2 ) = 26/50 (26 cards not in AS or AH piles) P(E 4 E 1 E 2 E 3 ) = 13/49 (13 cards not in AS, AH, AD piles)

12 piling cards E 1 = { in any one pile } E 2 = { and in different piles } E 3 = { different piles } E 4 = { all four aces in different piles } P(E 1 E 2 E 3 E 4 ) = P(E 1 ) P(E 2 E 1 ) P(E 3 E 1 E 2 ) P(E 4 E 1 E 2 E 3 ) = ( )/( ) 0.105

13 law of total probability E and F are events in the sample space S E = EF EF c S E F EF EF c = P(E) = P(EF) + P(EF c )

14 law of total probability P(E) = P(EF) + P(EF c ) = P(E F) P(F) + P(E F c ) P(F c ) = P(E F) P(F) + P(E F c ) (1-P(F)) weighted average, conditioned on event happening or not. More generally, if F1, F2,..., Fn partition S (mutually exclusive, i Fi = S, P(Fi)>0), then P(E) = i P(E Fi) P(Fi)

15 Bayes Theorem Most common form: P(F E) = P(EF)/P(E) = [P(E F) P(F)]/P(E) posterior vs prior; reverse conditioning Expanded form (using law of total probability):

16 Improbable Inspiration: The future of software may lie in the obscure theories of an 18 th century cleric named Thomas Bayes Los Angeles Times (October 28, 1996) By Leslie Helm, Times Staff Writer Bayes Theorem When Microsoft Senior Vice President Steve Ballmer [now CEO] first heard his company was planning a huge investment in an Internet service offering he went to Chairman Bill Gates with his concerns Gates began discussing the critical role of Bayesian systems source:

17 HIV testing Suppose an HIV test is 98% effective in detecting HIV, i.e., its false negative rate = 2%. Suppose furthermore, the test s false positive rate = 1%. 0.5% of population has HIV Let E = you test positive for HIV Let F = you actually have HIV What is P(F E)? Solution:

18 HIV+ why it s still good to get tested HIV- Test = P(E F) 0.01 = P(E F c ) Test = P(E c F) 0.99 = P(E c F c ) Let E c = you test negative for HIV Let F = you actually have HIV What is P(F E c )?

19 simple spam detection Say that 60% of is spam 90% of spam has a forged header 20% of non-spam has a forged header Let F = message contains a forged header Let J = message is spam What is P(J F)? Solution:

20 simple spam detection Say that 60% of is spam 10% of spam has the word Viagra 1% of non-spam has the word Viagra Let V = message contains the word Viagra Let J = message is spam What is P(J V)? Solution:

21 Child is born with (A,a) gene pair (event B A,a ) Mother has (A,A) gene pair Two possible fathers: M 1 = (a,a), M 2 = (a,a) P(M 1 ) = p, P(M 2 ) = 1-p What is P(M 1 B A,a )? Solution: DNA paternity testing

22 odds The odds of event E is P(E)/(P(E c ) Example: A = any of 2 coin flips is H: P(A) = 3/4, P(A c ) = 1/4, so odds of A is 3 (or 3 to 1 in favor ) Example: odds of having HIV: P(F) =.5% so P(F)/P(F c ) =.005/.995 (or 1 to 199 against)

23 posterior odds from prior odds F = some event of interest (say, HIV+ ) E = additional evidence (say, HIV test was positive ) Prior odds of F: P(F)/P(F c ) What are the Posterior odds of F: P(F E)/P(F c E)?

24 Let E = you test positive for HIV Let F = you actually have HIV What are the posterior odds? posterior odds from prior odds More likely to test positive if you are positive, so Bayes factor >1; positive test increases odds 98-fold, to 2.03:1 against (vs prior of 199:1 against)

25 posterior odds from prior odds Let E = you test negative for HIV Let F = you actually have HIV What is the ratio between P(F E) and P(F c E)? Unlikely to test negative if you are positive, so Bayes factor <1; negative test decreases odds 49.5-fold, to 9850:1 against (vs prior of 199:1 against)

26 simple spam detection Say that 60% of is spam 10% of spam has the word Viagra 1% of non-spam has the word Viagra Let V = message contains the word Viagra Let J = message is spam What are posterior odds that a message containing Viagra is spam? Solution:

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