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K 1 i 1 a i n k 1 x i 1 p a i n k 1 p a 1 where the

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- k +1 i =1 A i ) N - k +1 X i =1 P ( A i ) = ( N - k + 1) P ( A 1 ) , where the last equality follows from the fact that the N - k + 1 events A i have the same probability and this in turn follows from the assumption that the N 2
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bytes are equally likely to take any value in { 0 ,..., 255 } . Now P ( A i ) = ( 1 256 ) k , consequently P ( N - k +1 i =1 A i ) N - k +1 X i =1 P ( A i ) = N - k + 1 256 k . For k = 4 and N = 1024 , the probability of having k consecutive bytes some- where on the hard drive is upper bounded by 2 . 38 × 10 - 7 . For k = 4 and N = 1024 2 , the probability of having k consecutive bytes some- where on the hard drive is upper bounded by 2 . 44 × 10 - 4 . 6. Uncountable Ω and the Third Axiom The Harvard student’s mistake lies in the equality P ( x [0 , 1] { x } ) = Z 1 0 P ( { x } ) dx. For this, the Harvard student seems to have the second axiom in mind (using the book’s numbering), as s/he is expressing the probability of a disjoint union as a “sum” (actually, an integral) of the probabilities of the constituent events. But this is incorrect because the second axiom of probability holds only for a countable union of events: if A 1 ,A 2 ,... are disjoint events then P ( i =1 A i ) = i =1 P ( A i ) . In this case the set { x : x [0 , 1] } is uncountable and as such the second axiom does not apply here. 7. Manufacturing Defects Consider the square S whose center is the center of the transistor and with side length a + b . If the center of the dust particle falls anywhere inside S then it will touch the transistor and as a result the transistor will stop functioning. Thus we
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k 1 i 1 A i N k 1 X i 1 P A i N k 1 P A 1 where the last...

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