Final Exam - Mathematics Department, UCLA T. Richthammer...

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Mathematics Department, UCLA T. Richthammer winter 09, final Mar 02, 2009 Final: Math 170A Probability, Sec. 1 1. (6 pts) Consider a probability model with S = { 1 , 2 , 3 , 4 } , P ( { 1 , 2 } ) = 1 2 , P ( { 1 , 3 } ) = 1 4 . (a) Is it possible that P ( { 4 } ) = 0? (b) What are the possible values of P ( { 4 } )? Answer: Let p i := P ( { i } ). We have p 1 + p 2 = 1 2 , p 1 + p 3 = 1 4 and p 1 + p 2 + p 3 + p 4 = 1. (a) If p 4 = 0, then from above we get p 2 = 1 - 1 4 = 3 4 > p 1 + p 2 , which is not possible. (b) For given p 4 we have : p 3 = 1 - ( p 1 + p 2 ) - p 4 = 1 2 - p 4 , p 1 = ( p 1 + p 3 ) - p 3 = 1 4 - 1 2 + p 4 = p 4 - 1 4 and p 2 = ( p 1 + p 2 ) - p 1 = 1 2 - p 4 + 1 4 = 3 4 - p 4 . These probabilities are all in [0 , 1] for p 4 [ 1 4 , 1 2 ], so these are the possible values for p 4 . 2. (10 pts) Define probability models for the following situations, and give a precise description of the event A . ( Don’t calculate P(A)!) (a) A fair die is rolled 4 times. A = “The sum of the values is at most 10”. (b) A point is chosen within a cube of side length 8 completely at random. A = ”The distance of the point to the surface of the cube is less than 1“. (c) A number is chosen completely at random from { 1 , . . . , n } . If this number is k , then a fair coin is tossed k times and the number of heads is recorded. A = ”Head appears in more the half of the coin tosses”. Answer: (a) S = { 1 , . . . , 6 } 4 , P ( A ) = | A | / | S | . The projections X 1 , . . . , X 4 denote the outcomes of the 1st, . . . 4th roll. A = { X 1 + . . . + X 4 = 10 } . (b) S = [0 , 8] 3 , P ( A ) = λ 3 ( A ) 3 ( S ), where λ 3 denotes volume. A = S - [1 , 7] 3 . (c) S = { 1 , . . . , n } 2 . The first projection X 1 denotes the number chosen, the second projection X 2 the number of heads. P (( k, l )) = p k p l | k , where p k = 1 n and p l | k = ( k l ) 1 2 k (number of successes in k independent trials). A = { X 2 > X 1 / 2 } . 3. (9 pts) Suppose that 8 candies are distributed to 8 persons. Determine the probability that everybody gets a candy, assuming that all possible distributions are equally likely and (a) the candies are distinct (b) the candies are identical. In both cases define a model. You don’t have to simplify your results, but determine in which case the probability is higher. Answer: (a) S = { 1 , . . . , 8 } 8 = { ( x 1 , . . . , x 8 ) : x i ∈ { 1 , . . . , 8 }} , where x i denotes the person that gets candy i . P ( A ) = | A | / | S | . Here | S | = 8 8 , | A | = 8!, so P ( A ) = 8! / 8 8 . (b) S = { ( x 1 , . . . , x 8 ) : x i ∈ { 0 , 1 , 2 . . . } , x 1 + . . . x 8 = 8 } , where x i denotes the number of candies for person i . P ( A ) = | A | / | S | . Here | S | = ( 15 7 ) , | A | = 1, so P ( A ) = 7!8!
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This note was uploaded on 07/07/2010 for the course MATH 170A 170A taught by Professor Richthammer during the Winter '10 term at UCLA.

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Final Exam - Mathematics Department, UCLA T. Richthammer...

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