mid1-2010soln2 - 332:226 Probability and Stochastic...

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332:226 Probability and Stochastic Processes Examination 1 March 10, 2010 SOLUTION You have 110 minutes to answer the following four questions in the notebooks provided. This is a closed book exam; neither notes nor calculators are permitted . Make sure that you have included your name, Rutgers netid and signature in each book used (5 points). Leave your exam paper stapled to your exam blue book (5 points). Read each question carefully. All statements must be justified. Computations should be simplified as much as possible. 1. 70 points SHORT ANSWERS: You must explain your answers to these unrelated questions. (a) In a game of poker, you are dealt a five card hand. What is the probability P [ R ] that your hand has only red cards? One way to solve this problem is to observe that There are ( 52 5 ) equally likely five-card hands. There are ( 26 5 ) five-card hands of all red cards. Thus the probability getting a hand of all red cards is P [ R ] = ( 26 5 ) ( 52 5 ) . Note that this can be rewritten as P [ R ] = 26 52 25 51 24 50 23 49 22 48 , where the fractions are the conditional probabilities of receiving successive red cards. (b) For events A and B , P [ A B ] = P [ A ] - P [ B ]. What is P [ B ]? P [ A ] P [ A B ] = P [ A ] - P [ B ] , P [ B ] 0 , P [ B ] = 0 . (c) X is a discrete uniform (1 , 5) random variable. What is P [ X > E [ X ]]? P X ( x ) = ( 1 / 5 x = 1 , 2 , 3 , 4 , 5 , 0 otherwise . E [ X ] = 5 X x =1 xP X ( x ) = 1 5 5 X x =1 x = 3 . P [ X > E [ X ]] = P [ X > 3] = P X (4) + P X (5) = 2 5 . (d) Y is a continuous uniform (1 , 5) random variable. What is P [ Y > E [ Y ]]? f Y ( y ) = ( 1 / 4 1 x 5 , 0 otherwise . E [ Y ] = Z 5 1 y 4 dy = y 2 8 5 1 = 3 . P [ Y > E [ Y ]] = P [ Y > 3] = Z 5 3 1 4 dy = 1 2 . 1
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(e) In a bag of 25 M&Ms, each piece is equally likely to be red, green, orange, blue, or brown, independent of the color of any other piece. Find the the PMF of R , the number of red pieces.
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