Practice_Midterm(2)_Sol_STAT311

# Practice_Midterm(2)_Sol_STAT311 - STAT 311 Practice...

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STAT 311 Practice Questions Q1. We are drawing balls from the urn with 2 white balls and 3 red balls with replacement. X be the number of red balls drawn in the first two draw, then X=0, 1, 2. Y can go on forever until you get the first red ball in your draw. (That means, Y has a geometric distribution, and Y=1,2,3, ..) (a) Joint probability mass function of p X,Y (x, y) In this kind of example, (When things do not look that straightforward), trying to figure out the exact kind of distribution may be rather futile. Instead calmly think about all the possible cases! First fixate on X. When X=0 what would Y be? Then when X=1, what would be the corresponding Y? Then when X=2? Joint probability mass function is what assigns the probability to each possible pair of X and Y. Remember, the value of what Y can take is restricted with the condition, first red ball is drawn , therefore, bounded by the value of X. - (X=0, Y=1): No such thing can happen - (X=0, Y=2): No such thing can happen - (X=0, Y>2): White balls are drawn in the first two draw, and then whatever happens after that does not matter. (WW) - (X=1, Y=1): When the drawn balls are RW - (X=1, Y=2): When the drawn balls are WR - (X=1, Y>2): No such thing can happen - (X=2, Y=1): When the drawn balls are RR - (X=2, Y=2): No such thing can happen - (X=2, Y>2): No such thing can happen P X = 0, Y = 1 = 0 P X = 0, Y = 2 = 0 P X = 0, Y > 2 = 2 5 2 5 = 4 25 P X = 1, Y = 1 = 3 5 2 5 = 6 25 P X = 1, Y = 2 = 2 5 3 5 = 6 25 P X = 1, Y > 2 = 0 P X = 2, Y = 1 = 3 5 3 5 = 9 25 P X = 2, Y = 2 = 0 P X = 2, Y > 2 = 0 To make sure, add up all the probabilities and see if it is equal to 1. (b) Probability mass function of X and Y assigns probabilities to the possible values of X and Y P X = 0) = P(X = 0, Y = 1 + P X = 0, Y = 2 + P X = 0, Y > 2 = 0 + 0 + 4 25 = 4 25 P X = 1) = P(X = 1, Y = 1 + P X = 1, Y = 2 + P X = 1, Y > 2 = 0 + 6 25 + 6 25 = 12 25 P X = 2) = P(X = 2, Y = 1 + P X = 2, Y = 2 + P X = 2, Y > 2 = 9 25 + 0 + 0 = 9 25 Again, make sure that the sum is equal to 1 P Y = 1) = P(X = 0, Y = 1 + P X = 1, Y = 1 + P X = 2, Y = 1 = 0 + 6 25 + 9 25 = 15 25

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P Y = 2) = P(X = 0, Y = 2 + P X = 1, Y = 2 + P X = 2, Y = 2 = 0 + 6 25 + 0 = 6 25 P Y > 2) = P(X = 0, Y > 2 + P X = 1, Y > 2 + P X = 2, Y > 2 = 4 25 + 0 + 0 = 4 25 Again, make sure that the sum is equal to 1 This is actually, exactly the geometric case, since the experiment goes on until you get a red ball, and the probability of getting a red ball is 3 5 . P Y = y = 1 3 5 y 1 ( 3 5 ) , y = 1,2, To verify it with our result, P Y = 1 = 3 5 , P Y = 2 = 6 25 , P Y > 2 = 2 5 y 1 3 5 = 2 5 2 3 5 1 2 5 y=3 = 4 25 . (c) I {Y=2} = 1 when Y = 2, but otherwise = 0 X I {Y=2} : First forget about E() and think of it case-by-case (X=0, Y=1): I {Y=2} = 0, and likewise, X I Y=2 = 0 (X=0, Y=2): I {Y=2} = 1, and therefore, X I {Y=2} = X(= 0) (X=0, Y>2): I {Y=2} = 0 and therefore, X I {Y=2} = 0 (X=1, Y=1): I {Y=2} = 0, and likewise, X I Y=2 = 0 (X=1, Y=2): I {Y=2} = 1, and therefore, X I {Y=2} = X(= 1) (X=1, Y>2): I {Y=2} = 0 and therefore, X I {Y=2} = 0 (X=2, Y=1): I {Y=2} = 0, and likewise, X I Y=2 = 0 (X=2, Y=2): I {Y=2} = 1, and therefore, X I {Y=2} = X(= 2) (X=2, Y>2): I {Y=2}
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