Because the sample space is finite we only need to

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Because the sample space is finite, we only need to assign probabilities to the outcomes. For a fair coin, it is appropriate for all outcomes to be equally likely. That is, we use the uniform probability measure. (b) The probability that at least one flip comes up heads is given by P ( At least one flip comes up heads ) = | At least one heads | | Ω | = 7 8 , where | A | designates the cardinality of A or the number of its elements. 4. (a) Let A 1 , ...A n be a collection of events (not necessarily disjoint) such that P ( A i ) = 0 for all i . From the union bound we obtain P ( n i =1 A i ) n X i =1 P ( A i ) = n X i =1 0 = 0 . Since P ( A ) 0 for all A Ω , we deduce that P ( n i =1 A i ) = 0 . (b) Let A 1 , ...A n be a collection of events such that P ( A i ) = 1 for all i . P ( n i =1 A i ) = 1 - P (( n i =1 A i ) c ) = 1 - P ( n i =1 A c i ) . However we have P ( A c i ) = 1 - P ( A i ) = 0 , hence from (a) we deduce that P ( n i =1 A c i ) = 0 , i.e., P ( n i =1 A i ) = 1 . 5. The Union Bound Let A i for i ∈ { 1 , 2 , ..., N - k +1 } be the event that the k consecutive bytes start from the i th byte, for instance A 1 = { 0 ... 0 | {z } k times b k +1 ...b N } , where b j are arbitrary bytes. Then N - k +1 i =1 A i represents the event that somewhere on the hard drive there are k consecutive zeros. From the union bound we have P ( N - 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 .
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