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For k 4 and n 1024 the probability of having k

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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 have P ( Transistor does not work ) = Area of the square S Area of the outer square = ( a + b ) 2 c 2 . Consequently, we deduce that P ( Transistor works ) = 1 - P ( Transistor does not work ) = 1 - ( a + b ) 2 c 2 . 8. Waiting for Patterns The following MATLAB code will simulate a sequence of coin flips and return the number of flips it takes to see that pattern HT. 3
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function flipcount = findHT flipcount = 0; ht_count = 0; while (ht_count < 2), flip = round(rand); flipcount = flipcount + 1; % ht_count = 0; there hasn’t been a head so far if (ht_count == 0), % if heads, increment ht_count if (flip == 1),
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