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lecture03 - ECE 5510 Random Processes Lecture Notes Fall...

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ECE 5510: Random Processes Lecture Notes Fall 2009 Lecture 3 Today: (1) Conditional Probability, (2) Trees, (3) Total Prob- ability Announcements: 1. HW 1 due Thursday at 5pm. It is a short HW, don’t get behind. 2. Reading for today: Sections 1.5,1.7, Mlodinow handout. 3. Reading for Thursday: 1.8-1.9, and 2.1-2.4. 1 Conditional Probability Example: Three Card Monte (Credited to Prof. Andrew Yagle, U. of Michigan.) There are three two-sided cards: red/red, red/yellow, yellow/yellow. The cards are mixed up and shuffled, one is selected at random, and you look at one side of that card. You see red. What is the prob- ability that the other side is red? Three possible lines of reasoning on this: 1. Bottom card is red only if you chose the red/red card: P = 1 / 3. 2. You didn’t pick the yellow/yellow card, so either the red/red card or the red/yellow card: P = 1 / 2. 3. There are five sides which we can’t see, two red and three yellow: P = 2 / 5. Which is correct? Def’n: Conditional Probability, P [ A | B ] P [event A occurs, GIVEN THAT event B occurred] For events A and B , when P [ B ] > 0, P [ A | B ] defines P [ A B ] P [ B ] = P [ A B ] P [ A B ] + P [ A c B ] Notes:
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ECE 5510 Fall 2009 2 1. Given that B occurs, now we know that either A B occurs, or A c B occurs. 2. We’re defining a new probability model, knowing more about the world. Instead of P [ · ], we call this model P [ ·| B ]. All of our Axioms STILL APPLY!
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