Flip a fair coin n times. using markov's and
chebyshev'sinequalities ,prove that the length of the longest
constant run inthe resulting list of heads and tails is (1 +
o(1))log_{2}(n). inother words, for every a> 0, almost no list
has at least(1+a)log_{2}(n) consecutive identical flips, and almost
every listhas at least (1-a)log_{2}(n) consecutive identical
flips.

see explanation

step-by-step explanation:

since the sequence is geometric there is a common ratio r between consecutive terms, that is

r = 4 ÷ - 2 = - 2

r = - 8 ÷ 4 = - 2

r = 16 ÷ 8 = - 2

the next term - 13 must be incorrect as it should be - 32 for the sequence to be geometric.

the first 4 terms are therefore geometric with common ratio r = - 2

9987

step-by-step explanation:

a

step-by-step explanation:

the denominator cannot be the same as one of the factors.

8 denominator times 2 denominator = 16

and 7 numerator time 11 = 77

77/16= 77/16 ( they cannot be further simplified)

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