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# 4 - It follows that pX(k =(111 k)3(110 k)3 103 0 if k = 101...

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It follows that p X ( k ) = (111 k ) 3 (110 k ) 3 10 3 , if k = 101 , . . . 110, 0 , otherwise. (An alternative solution is based on the notion of a CDF, which will be introduced in Chapter 3.) (b) Since X i is uniformly distributed over the integers in the range [101 , 110], we have E [ X i ] = (101 + 110) / 2 = 105 . 5. The expected value of X is E [ X ] = k = −∞ k · p X ( k ) = 110 k =101 k · p x ( k ) = 110 k =101 k · (111 k ) 3 (110 k ) 3 10 3 . The above expression can be evaluated to be equal to 103.025. The expected improve- ment is therefore 105.5 - 103.025 = 2.475. Solution to Problem 2.31. The marginal PMF p Y is given by the binomial formula p Y ( y ) = 4 y 1 6 y 5 6 4 y , y = 0 , 1 , . . . , 4 . To compute the conditional PMF p X | Y , note that given that Y = y , X is the number of 1’s in the remaining 4 y rolls, each of which can take the 5 values 1 , 3 , 4 , 5 , 6 with equal probability 1/5. Thus, the conditional PMF p X | Y is binomial with parameters 4 y and p = 1 / 5: p X | Y ( x | y ) = 4 y x 1 5 x 4 5 4 y x , for all nonnegative integers x and y such that 0 x + y 4. The joint PMF is now given by p X,Y ( x, y ) = p Y ( y ) p X | Y ( x | y ) = 4 y 1 6 y 5 6 4 y 4 y x 1 5 x 4 5 4 y x , for all nonnegative integers x and y such that 0 x + y 4. For other values of x and y , we have p X,Y ( x, y ) = 0. Solution to Problem 2.32. Let X i be the random variable taking the value 1 or 0 depending on whether the first partner of the i th couple has survived or not. Let Y i be the corresponding random variable for the second partner of the i th couple. Then, we have S = m i =1 X i Y i , and by using the total expectation theorem, E [ S | A = a ] = m i =1 E [ X i Y i | A = a ] = m E [ X 1 Y 1 | A = a ] = m E [ Y 1 = 1 | X 1 = 1 , A = a ] P ( X 1 = 1 | A = a ) = m P ( Y 1 = 1 | X 1 = 1 , A = a ) P ( X 1 = 1 | A = a ) . 31

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We have P ( Y 1 = 1 | X 1 = 1 , A = a ) = a 1 2 m 1 , P ( X 1 = 1 | A = a ) = a 2 m . Thus E [ S | A = a ] = m a 1 2 m 1 · a 2 m = a ( a 1) 2(2 m 1) . Note that E [ S | A = a ] does not depend on p . Solution to Problem 2.38. (a) Let X be the number of red lights that Alice encounters. The PMF of X is binomial with n = 4 and p = 1 / 2. The mean and the variance of X are E [ X ] = np = 2 and var( X ) = np (1 p ) = 4 · (1 / 2) · (1 / 2) = 1. (b) The variance of Alice’s commuting time is the same as the variance of the time by which Alice is delayed by the red lights. This is equal to the variance of 2 X , which is 4var( X ) = 4. Solution to Problem 2.39. Let X i be the number of eggs Harry eats on day i . Then, the X i are independent random variables, uniformly distributed over the set { 1 , . . . , 6 } . We have X = 10 i =1 X i , and E [ X ] = E 10 i =1 X i = 10 i =1 E [ X i ] = 35 . Similarly, we have var( X ) = var 10 i =1 X i = 10 i =1 var( X i ) , since the X i are independent. Using the formula of Example 2.6, we have var( X i ) = (6 1)(6 1 + 2) 12 2 . 9167 , so that var( X ) 29 . 167. Solution to Problem 2.40. Associate a success with a paper that receives a grade that has not been received before. Let X i be the number of papers between the i th success and the ( i + 1)st success. Then we have X = 1 + 5 i =1 X i and hence E [ X ] = 1 + 5 i =1 E [ X i ] .
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4 - It follows that pX(k =(111 k)3(110 k)3 103 0 if k = 101...

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