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seen Aug 29 '13 at 14:05

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