elemprob-fall2010-page23

elemprob-fall2010-page23 - or P(F ∩ E ) = P(E )P(F ),...

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Unformatted text preview: or P(F ∩ E ) = P(E )P(F ), which agrees with the definition of independence we gave before. Let us give two more examples. An example: Suppose an urn holds 5 red balls and 7 green balls. You draw two balls without replacement. What is the probability the second ball is red? Answer. Let A be the event that the first ball is red, B that the second ball is, and we want P(B ). Then P(B ) = P(A ∩ B ) + P(Ac ∩ B ) = P(B | A)P(A) + P(B | Ac )P(Ac ). 5 7 The probability of A is 12 and the probability for Ac is 12 . Given that the 4 first ball is red, there are now 4 red balls and 7 green, so P(B | A) = 11 . 5 Similarly, P(B | Ac ) = 11 . Therefore P(B ) = 57 5 45 · + · =, 11 12 11 12 12 which is what one would expect. An example. This is known as the Monty Hall problem after the host of the TV show of the 60’s called Let’s Make a Deal. There are three doors, behind one a nice car, behind each of the other two a bale of straw. You choose a door. Then Monty Hall opens one of the other doors, which shows a bale of straw. He gives you the opportunity of switching to the remaining door. Should you do it? Answer. Let’s suppose you choose door 1, since the same analysis applies whichever door you chose. Strategy one is to stick with door 1. With probability 1/3 you chose the car. Monty Hall shows you one of the other doors. Since you know he won’t show you the door with the car, your probability is still 1/3. Strategy 2 is to change. With probability 1/3 you chose the car. He shows you another door, say, door 2, with straw, and then you choose door 3 and lose. With probability 2/3 the car is behind door 2 or 3. He shows you the door the car is not behind, say, door 2, and you change to door 3, and win. So switching increases your probability of winning to 2/3. 23 ...
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This note was uploaded on 12/29/2011 for the course MATH 316 taught by Professor Ansan during the Spring '10 term at SUNY Stony Brook.

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