cse 321 discrete structures winter 2008 lecture 19 probability theory texpoint fonts used in emf....

CSE 321 Discrete Structures

Winter 2008

Lecture 19

Probability Theory

Announcements

• Readings – Probability Theory

• 6.1, 6.2 (5.1, 5.2) Probability Theory• 6.3 (New material!) Bayes’ Theorem• 6.4 (5.3) Expectation

– Advanced Counting Techniques – Ch 7.• Not covered

Discrete Probability

Experiment: Procedure that yields an outcome

Sample space: Set of all possible outcomes

Event: subset of the sample space

S a sample space of equally likely outcomes, E an event, the probability of E, p(E) = |E|/|S|

Example: Dice

Example: Poker

Probability of 4 of a kind

Combinations of Events

EC is the complement of E

P(EC) = 1 – P(E)

P(E1 E2) = P(E1) + P(E2) – P(E1 E2)

Combinations of Events

EC is the complement of E

P(EC) = 1 – P(E)

P(E1 E2) = P(E1) + P(E2) – P(E1 E2)

Probability Concepts

• Probability Distribution

• Conditional Probability

• Independence

• Bernoulli Trials / Binomial Distribution

• Random Variable

Discrete Probability Theory

• Set S

• Probability distribution p : S [0,1]– For s S, 0 p(s) 1

– sS p(s) = 1

• Event E, E S

• p(E) = sEp(s)

Examples

Conditional Probability

Let E and F be events with p(F) > 0. The conditional probability of E given F, defined by p(E | F), is defined as:

Examples

Independence

The events E and F are independent if and only if p(E F) = p(E)p(F)

E and F are independent if and only if p(E | F) = p(E)

Are these independent?

• Flip a coin three times– E: the first coin is a head– F: the second coin is a head

• Roll two dice– E: the sum of the two dice is 5– F: the first die is a 1

• Roll two dice– E: the sum of the two dice is 7– F: the first die is a 1

• Deal two five card poker hands– E: hand one has four of a kind– F: hand two has four of a kind

Bernoulli Trials and Binomial Distribution

• Bernoulli Trial– Success probability p, failure probability q

The probability of exactly k successes in n independent Bernoulli trials is

Random Variables

A random variable is a function from a sample space to the real numbers

Bayes’ Theorem

Suppose that E and F are events from a sample space S such that p(E) > 0 and p(F) > 0. Then

False Positives, False Negatives

Let D be the event that a person has the disease

Let Y be the event that a person tests positive for the disease

Testing for disease

Disease is very rare: p(D) = 1/100,000

Testing is accurate:False negative: 1%False positive: 0.5%

Suppose you get a positive result, whatdo you conclude?

P(D|Y)

Spam Filtering

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Bayesian Spam filters

• Classification domain– Cost of false negative– Cost of false positive

• Criteria for spam– v1agra, ONE HUNDRED MILLION USD

• Basic question: given an email message, based on spam criteria, what is the probability it is spam

Email message with phrase “Account Review”

• 250 of 20000 messages known to be spam• 5 of 10000 messages known not to be spam• Assuming 50% of messages are spam, what

is the probability that a message with “Account Review” is spam

Proving Bayes’ Theorem

Expectation

The expected value of random variable X(s) on sample space S is:

Flip a coin until the first headExpected number of flips?

Probability Space:

Computing the expectation:

Linearity of Expectation

E(X1 + X2) = E(X1) + E(X2)

E(aX) = aE(X)

Hashing

H: M [0..n-1]

If k elements have been hashed torandom locations, what is the expected number of elements in bucket j?

What is the expected number ofcollisions when hashing k elements to random locations?

Hashing analysis

Sample space: [0..n-1] [0..n-1] . . . [0..n-1]

Random VariablesXj = number of elements hashed to bucket jC = total number of collisions

Bij = 1 if element i hashed to bucket jBij = 0 if element i is not hashed to bucket j

Cab = 1 if element a is hashed to the same bucket as element bCab = 0 if element a is hashed to a different bucket than element b

Counting inversions

Let p1, p2, . . . , pn be a permutation of 1 . . . npi, pj is an inversion if i < j and pi > pj

4, 2, 5, 1, 3

1, 6, 4, 3, 2, 5

7, 6, 5, 4, 3, 2, 1

Expected number of inversions for a random permutation

Insertion sort

for i :=1 to n-1{ j := i; while (j > 0 and A[ j - 1 ] > A[ j ]){

swap(A[ j -1], A[ j ]); j := j – 1; }}

4 2 5 1 3

Expected number of swaps for Insertion Sort

Left to right maxima

max_so_far := A[0];for i := 1 to n-1 if (A[ i ] > max_so_far)

max_so_far := A[ i ];

5, 2, 9, 14, 11, 18, 7, 16, 1, 20, 3, 19, 10, 15, 4, 6, 17, 12, 8

What is the expected number of left-to-right maxima in a random permutation