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 Jul 22 awarded Yearling May 6 awarded Good Answer Aug 26 awarded Yearling Jun 23 comment Lower bounds on the probability that one random variable is greater than a set of others @Emre, no they are not identically distributed Jun 23 asked Lower bounds on the probability that one random variable is greater than a set of others Dec 22 comment Can we partition NP-complete problem into finite number of polynomially solvable problems? @Oleksandr While, running a finite number of polynomial time algorithms requires only polynomial time we don't know which answer is correct. For example, we can partition all instances of 3-Sat into two sets: YES and NO, where YES contains all instances that are satsifiable and NO contains all instances that are unsatisfiable. Trivially, all problems in YES can be solved in polynomial time by the Turing Machine (TM) that just accepts and all problems in NO can be solved in polynomial time by the TM that just rejects. However, this tells us nothing about whether a given instance is satisfiable. Dec 21 comment In a family with two children, what are the chances, if one of the children is a girl, that both children are girls? @Nemi You are correct that there is an important distinction between "one of the two children is a girl" and "the first child is a girl". My text says: "Since we are given that at least one child is a girl there are three possibilities: bg, gb, or gg". I did not say that we are given that the first child is a girl. The three possibilities arise from considering whether the first/second born was a girl/boy. Dec 21 awarded Nice Answer Dec 21 awarded Teacher Dec 21 answered In a family with two children, what are the chances, if one of the children is a girl, that both children are girls? Dec 16 awarded Scholar Dec 16 accepted Simplify $\sum \limits_{k=0}^{n} \binom{n}{k} 2^{\sqrt{k}}$ Dec 16 comment Simplify $\sum \limits_{k=0}^{n} \binom{n}{k} 2^{\sqrt{k}}$ @Shai Covo yes assume $a,b \geq 1$. In fact I am also interested in the case were $a=b \geq 1$ Dec 16 comment Simplify $\sum \limits_{k=0}^{n} \binom{n}{k} 2^{\sqrt{k}}$ I care more about asymptotic bounds so the floor or the full irrational value is fine with me Dec 15 awarded Supporter Dec 15 awarded Editor Dec 15 revised Simplify $\sum \limits_{k=0}^{n} \binom{n}{k} 2^{\sqrt{k}}$ added 763 characters in body Dec 15 awarded Student Dec 15 asked Simplify $\sum \limits_{k=0}^{n} \binom{n}{k} 2^{\sqrt{k}}$ Aug 25 answered Estimates involving sums with binomials