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A level of a max heap (containing 100 nos) is choosen randomly, on its selection, a node from the same level is choosen randomly. What is the probability that it is the 36th  smallest element
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what I did is

P = $\frac{1}{7}*0 + \frac{1}{7}*\frac{1}{2} + \frac{1}{7}*\frac{1}{4} + \frac{1}{7}*\frac{1}{8}+ \frac{1}{7}*\frac{1}{16}+ \frac{1}{7}*\frac{1}{32}+ \frac{1}{7}*\frac{1}{37} = 0.087$

Suggestion and correction plz !
@Debashish Deka u r correct...

having one dout ... 36 at level 2 is at fixed location at right end and having only one choice..here u takes probability=1/2 since this level having 2 ele.

now at level 3 having 4 elements ...but at this level 36 can be at any place ... u takes probability=1/4 here having 4 choices...

confusing probability for me! plz explain !
@Debashish , why 36 occurs in any level from 2 to 7.
@debashish, can you make that an answer

Its P(65th largest element given that a level is chsen and an elemwnt is chosen from it)

So P= ((1/7)*(1/37))/((1/7)+(1/7)*(1/2)+......+ (1/7)*(1/2^5)+(1/7)*(1/37))

P=0.0135