GEOMETRIC DISTRIBUTIONS

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Observations Theoretically, this distribution couldcontinue forever, since one possibleoutcome is that the player never rollsdoubles. However, the probabilitiesdecrease considerably as the waiting timeincreases. Since the sum of P(X) must equal1, this distribution becomes an infinitegeometric series.

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Probability Distribution Binomial Distribution Geometric Distribution ...

Solutiona) X = waiting time X = {0, 1, 2, ...} p = P(doubles) = 1/6 q = 5/6Unsuccessful rolls(Waiting Time), xProbabilityP(x)0 (5/6) 0 (1/6) = 0.1671 (5/6) 1 (1/6)=0.1392 (5/6) 2 (1/6)=0.1163 (5/6) 3 (1/6)=0.0964 (5/6) 4 (1/6)=0.080

Page 1 and 2: GEOMETRICDISTRIBUTIONS
Page 3 and 4: Geometric Distribution The above ex
Page 5: Example1. a) Calculate the probabil
Page 9 and 10: Observation (cont) The Expected Val
Page 12 and 13: Solutiona) X = waiting timeb)= {0,

GEOMETRIC DISTRIBUTIONS

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