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Probability 2

Probability 2. Professor Jim Ritcey EE 416 Please elaborate with your own sketches. Disclaimer. These notes are not complete, but they should help in organizing the class flow. Please augment these notes with your own sketches and math . You need to actively participate.

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Probability 2

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  1. Probability 2 Professor Jim Ritcey EE 416 Please elaborate with your own sketches

  2. Disclaimer • These notes are not complete, but they should help in organizing the class flow. • Please augment these notes with your own sketches and math. You need to actively participate. • It is virtually impossible to learn this from a verbal description or these ppt bullet points. You must create your own illustrations and actively solve problems.

  3. Conditional Probabilities • Given (S, E, P) with events E ={ A,B,C, … } • Pick two events A and B. Define • P( B|A) = P(AB)/P(A) only when P(A) >0 • This is the conditional probability of B given A • Draw a picture using Venn diagrams! • It is often easier to remember that • P(AB) = P(B|A)P(A), • Recall that AB = A cap B , the intersection

  4. Conditional Prob & Independence • 2 events are independent when P(AB) = P(A) P(B) • Under independence • P(A|B) = P(A) & P(B|A) = P(B) • Under independence, the condition provides no new information as it leaves the probability unchanged

  5. Ranking Example (MacKay) • Fred has two brothers Alf and Bob. • What is the probability that Fred is older than Bob { B < F } • We can ignore Alf and the sample space is • Outcomes { B<F, F<B } equally likely ½ by insufficient reason to assume otherwise • But what if we include Alf?

  6. Ranking Example (MacKay) • We can include Alf and the sample space is • All 3!=6 rankings of A,B,F. Write ABF = A<B<F • Outcomes { ABF, AFB, FAB, BAF, BFA, FBA } equally likely 1/6 by insufficient reason. • Then {B <F} = {ABF,BAF,BFA} = 3/6 =1/2 • Read { a,b,c} = {a} OR {b} OR {c} • Now Fred says he is older than Alf {A <F} has occurred. Find P( B<F|A<F )? Enumerate!

  7. Bayes Rule • Bayes Rule is simply the equality derived by • P(A|B)P(B) = P(AB) = P(BA) = P(B|A)P(A) • Or P(A|B) = P(B|A)P(A)/P(B), & P(B)>0, P(A)>0 • Critical tool for inference – working backwards by induction

  8. Classic Vendor Example

  9. Classic Vendor Example

  10. Classic Vendor Example

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