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03cond - EECS 501 CONDITIONAL PROBABILITY Fall 2001 Three...

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EECS 501 CONDITIONAL PROBABILITY Fall 2001 Three There are 3 cards: red/red; red/black; black/black. Card The cards are shuffled and one chosen at random. Monte The top of the card is red. Pr[bottom is red]=? 3 possible lines of reasoning to solve this problem: 1. Bottom is red only if chose red/red card Pr = 1 / 3. 2. Not black/black, so either red/black or red/red Pr = 1 / 2. 3. 5 hidden sides: 2 red and 3 black Pr = 2 / 5. Which is correct? They are ALL wrong! In fact, Pr = 2 / 3. DEF: Pr [ A | B ] = Pr [event A occurs, GIVEN THAT event B occurrED]. Either A occurs or A doesn’t occur, even if B occurred. Their relative probabilities (ratio) shouldn’t change, after restriction to B occurring ( A B and A 0 B ). Want: Pr [ A | B ] + Pr [ A 0 | B ] = 1 and P r [ A | B ] P r [ A 0 | B ] = P r [ A B ] P r [ A 0 B ] . Know: Pr [ A B ] + Pr [ A 0 B ] = Pr [ B ]. So just divide this by Pr [ B ]. THM: Pr [ A | B ] = Pr [ A B ] /Pr [ B ] = Pr [ A B ] / ( Pr [ A B ] + Pr [ A 0 B ]). NOTE: Forms: Pr [ A | B ] = x x + y and Pr [ A 0 | B ] = y x + y . Ratio x/y , add to one.
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