5.2 - STAT3000 Section 5.2: The Binomial Distributions...

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STAT3000 Section 5.2: The Binomial Distributions Recall that about 45% of people have type O blood. Suppose you select 3 people at random and check their blood type. What is the probability that all 3 have type O blood? What is the probability that 2 have type O blood? What is the probability that only 1 has type O blood? What is the probability that none have type O blood? S = {OOO, OON,ONO, NOO, ONN, NON, NNO, NNN} P(3 type O) = P(OOO) = (O.45) (O.45) (O.45) = 1(0.45 3 ) = 0.0911 P(2 type O) = P(OON or ONO or NOO) = 3(0.45)(O.45)(0.55) = 3(0.45 2 )(0.55 1 ) = 0.3341 P(1 type O) = P(ONN or NON or NNO) = 3(0.45)(0.55)(0.55) = 3(0.45 1 )(0.55 2 ) = 0.4084 P(no type O) = P(NNN) =1 (0.55) (0.55) (0.55) = 1(0.55 3 ) = 0.1664 83
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Binomial Distribution 1. There are a fixed number n of observations. 2. The n observations are all independent . That is, knowing the result of one observation tells you nothing about the other observations. 3. Each observation falls into one of just two
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This note was uploaded on 06/05/2011 for the course STAT 3000 taught by Professor Staff during the Spring '08 term at University of Georgia Athens.

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5.2 - STAT3000 Section 5.2: The Binomial Distributions...

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