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Toss $n$ coins. What is the probability that we get more than $$ n/2 + 2\sqrt{n\cdot \log(n)}. $$

I have to use Chernoff Bounds here. If I let

$X_i$ indicate whether coin $i$ comes up heads,

$X=$ the summation of $X_i$,

I found the $E(X)=n/2$.

I can't seem to apply the Chernoff bounds now.

The answer to the question is $1/n^{8/3}$.

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From $E[X]=n/2$, one could infer that you have been told (but have not felt that it is important to reveal this to us) that the coins are fair and the tosses are independent. If so, please edit the question to include this information. The Chernoff bound says that for all $t>0$, $$P\{X \geq k\}=\sum_{i=k}^n \binom{n}{i}2^{-n}\leq\sum_{i=0}^n\binom{n}{i}2^{-n}e^{t(i-k)}=2^{-n}e^{-tk}(1+e‌​^t)^n$$ which has minimum value at $t=k/(n-k)$. Can you take it from here? –  Dilip Sarwate Apr 15 '13 at 1:33
    
Sorry, that should be at $t = \ln(k/(n-k))$, not $t=k/(n-k)$ –  Dilip Sarwate Apr 15 '13 at 2:01

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