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# DM-Probability -Q20

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+1 vote

An algorithm takes a list of $\dpi{100} 2^n$ numbers $\dpi{100} [ a_1,a_2,...,a_{2^n} ]$ and replaces it with $\dpi{100} [ b_1,b_2,...,b_{2^{n-1} } ]$, where $\dpi{100} b_1 = max(a_1,a_2)$,   $\dpi{100} b_2 = max(a_3,a_4)$  and so on. Then it performs the same operation on the resulting list (replacing each pair of consecutive elements with their maximum), and it continues doing the same until there are only two elements left in the list.

For instance, if the initial list is $\dpi{100} [3, 7, 6, 8, 2, 1, 4, 5]$, then after the first run, it becomes $\dpi{100} [7, 8, 2, 5]$ and then $\dpi{100} [8, 5]$

Suppose that the elements of the initial list are the integers 1 through 64 in random order. What is the probability that the number 63 will appear in the final two-element list?

(A). $\dpi{100} \frac{1}{63}$

(B) $\dpi{100} \frac{1}{4}$

(C) $\dpi{100} \frac{62}{63}$

(D) $\dpi{100} \frac{32}{63}$

asked Jul 7, 2020
reshown Jul 10, 2020