1 / 14

Lecture 8

Lecture 8. 5.2 Discrete Probability. 5.2 Recap. Sample space: space of all possible outcomes. Event: subset of of S. p(s) : probability of element s of S: Probability of complement Disjoint events: Overlapping events: Three doors problem (see ex. 39 p. 362 + matlab demo). S. HHH.

emelda
Download Presentation

Lecture 8

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Lecture 8 5.2 Discrete Probability

  2. 5.2 Recap • Sample space: space of all possible outcomes. • Event: subset of of S. • p(s) : probability of element s of S: • Probability of complement • Disjoint events: • Overlapping events: • Three doors problem (see ex. 39 p. 362 + matlab demo)

  3. S HHH E HTT THT HHT THH TTH TTT HTH F 5.2 We flip a coin 3 times with Heads and Tails equal probability. There are 8 possibilities: |S| = 8 E: event that an odd number of tails occurs F: first flip comes up tails

  4. 5.2 S THT E HTT THT HHT THH E|F THH TTH TTT HTH TTT F HHH TTH F What is the probability of odd number of tails (i.e. of event E) if we know that the first flip was tails (i.e. F happened) ?  If we know that F happened the total number of possible outcomes shrinks to |F| F = THH, THT, TTH, TTT. Of those there are two that make event E happen: THH, TTT. P(E|F) = 1/2

  5. Example: What is the probability that a family with 2 kids has two boys, given that they have at least one boy? (all possibilities are equally likely). S: all possibilities: {BB, GB, BG, GG}. E: family has two boys: {BB}. F: family has at least one boy: {BB, GB, BG}. E F = {BB} p(E|F) = (1/4) / (3/4) = 1/3 GG F BG E BB GB 5.2 Theorem: If E and F are two events and p(F)>0, then the conditional probability of E given F is given by:

  6. GG F BG E BB GB 5.2 S S HTT E THT HHT THH TTH TTT HTH F HHH p(E) = ¼ P(E|F) = 1/3 P(E)=1/2 P(E|F) = 1/2 By knowing F has happened, I have changed the probability that E has happened: they are dependent By knowing F has happened, I have NOT changed the probability that E has happened: they are independent

  7. 5.2 Equivalent statement: The probability of the intersection of 2 events is the product of the probabilities of the separate events. Example: A family has 2 children (|S|=4). Is the event E that the family has children of both sexes independent from the event F that they have at most one boy? E: {BG, GB} F: {GG, BG, GB} E F: {BG, GB} dependent

  8. 5.2 What about a family with 3 kids ? |S| = 8 |E| = 8 – 2 = 6 (only BBB and GGG violate E). |F| =| {GGG, GGB, BGG, GBG} | = 4 =| {BGG, GBG, GGB}| = 3 independent ! (again: it’s hard to get any intuition for this).

  9. 5.2 Problem: We toss a coin 7 times, but the coin is biased with probability 2/3 heads. What is the probability that we find 4 heads in those 7 tosses? There are C(7,4) ways to generate a sequence with 4 heads and 3 tails. (think of a bit-string with 0’s and 1’s). Each one of them has probability (2/3)^4 x (1/3)^3. Total: probability: C(7,4) x (2/3)^4 x (1/3)^3 More general: What is the probability of finding k 1’s in a bit-string of length n, when the probability of finding a 1 is p?  Binomial distribution. (matlab demo)

  10. 5.2 Does it sum to 1? Recall: Use q = 1-p to prove the result.

  11. 5.2 Random Variables Many problems deal with real numbers rather than set memberships. E.g. What is the sum of the outcomes of 2 dice? How many times do we expect to 7 1 in bit-strings of length 12? We were able to answer those questions before, but we now introduce some formal machinery: Definition:A random variable X is a function (not a variable!) from sample space to the real numbers. It assigns a real number to each possible outcome. (and is not random!). Example: S={apple, pear, banana}  X(apple) = 1, X(pear) = 2, X(banana) = 3. X: the number of times heads comes up when we toss a coin 2 times: X(TT) = 0; X(HT) = 1; X(TH) = 1; X(HH) = 2;

  12. 5.2 Definition: Let P(X=r) be the probability that X takes a value r. A distribution of a random variable X on a sample space S is the set of all pairs (r,P(X=r)) for all r in X(S). Thus, we specify the distribution by providing all possible P(X=r). Example: X is the number of heads of 2 coin tosses: P(X=0) = ¼ P(X=1) = ½ P(X=2) = ¼

  13. Sometimes, it’s more efficient to compute expectations by clumping together all elements of S that result in the same value for X(s). generally true only true when all p(s) s.t. X(s)=r have equal probability. number of elements in S such that X(s)=r 5.3 Definition: The expected value of a random variable X(s) on the sample space S is given by:

  14. 5.3 Example: X is number of heads in 2 coin tosses. Expected value? E(X) = 0 x P(X=0) + 1 x P(X=1) + 2 x P(X=2) = 0 + ½ + 2x ¼ = 1. We expect on average that we see 1 head. Example: X is the value of the number that comes up on a die. Expected value? E(X) = 1 x 1/6 + 2 x 1/6 + 3 x 1/6 + 4 x 1/6 + 5 x 1/6 + 6 x 1/6 = 21/6 = 7/2. (matlab demo2)

More Related