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Math135_hw5_soln

# Math135_hw5_soln - Math 135 Fall 2011 HW 5 This homework is...

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Math 135, Fall 2011: HW 5 This homework is due at the beginning of class on Friday October 14th, 2011 . You are free to talk with each other and get help. However, you should write up your own solutions and understand everything that you write. Problem 1 (p.182 #4) . Suppose all the numbers in a list of 100 numbers are nonnegative, and that the average of the numbers in the list is 2. Prove that at most 25 of the numbers in the list are greater than 8. SOLUTION. We use Markov’s inequality. Let X be a number drawn at random from the list. We know E [ X ] = 2. Hence P ( X > 8) E [ X ] 8 1 4 so at most 1/4 of the 100 numbers, i.e. 25 of the numbers in the list, can be bigger than 8. Problem 2 (p.182 #6) . Let X be the number of spades in 7 cards dealt from a well-shuffled deck of 52 cards containing 13 spades. Find E [ X ]. SOLUTION. The probability distribution of X , P ( X = i ), where i is an integer between 0 and 13, can be computed using the hypergeometric distribution: P ( X = i ) = ( 13 i )( 39 13 - i ) ( 52 13 ) E [ X ] = 13 X i =0 iP ( X = i ) Alternatively, one can solve this with the method of indicator functions: let I j be the indicator function of the event A j = { jth card is a spade } . Note that P ( A j ) = 13 52 . If you are concerned about the fact that, say, the probability of the second card being a spade depends on the previous card, observe that you can condition: P ( A 2 ) = P ( A 2 | A 1 ) P ( A 1 ) + P ( A 2 | A c 1 ) P ( A c 1 ) = 12 51 13 52 + 13 51 39 52 = 13 52 (Indeed, this was a homework problem on a previous homework set). Now, if X is the total number of spades in the hand, X = 7 X j =1 I j E ( X ) = E 7 X j =1 I j = 7 X j =1 E [ I j ] = 7 X j =1 13 52 = 7 4 1

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Problem 3 (p.182 #7) . In a circuit containing n switches, the i th switch is closed with probability p i , i = 1 , · · · , n . Let X be the total number of switches that are closed. Is it possible to calculate E [ X ] without further assumptions? If so, what is E [ X ]? SOLUTION. It is not possible to know the probability distribution of X without further assumptions. However, we can still compute the expected value of X by using the method of indicator functions. Let A i = { Switch i is closed } . Let I A i be the indicator function of A i , so I A i = 1 if switch i is closed, and 0 if switch i is open. We note that X = n X i =1 I A i because X is simply the total number of switches that are closed. Hence by the addition rule for expectation, E [ X ] = E " n X i =1 I A i # = n X i =1 E [ I A i ] = n X i =1 [1 P ( A i ) + 0 P ( A c i )] = n X i =1 P ( A i ) Problem 4 (p.182#10) . Let A and B
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