Question 1042698
You seem to be using the version {{{P(X > (1+epsilon)*mu) <= exp(-epsilon^2/3*mu)}}} of the Chernoff bound, where {{{X = sum(X[i], i = 1,n)}}}, and each {{{X[i]}}} is Bernoulli.


You want {{{P(X > n/2) <= 0.001}}}.

Let {{{(1+epsilon)mu = n/2}}}.   Since {{{mu = n/3}}}, we get

{{{(1+epsilon)(n/3) = n/2}}}.

===> {{{epsilon = 1/2}}}.

Next, let {{{exp(-epsilon^2/3*mu) = exp(-(1/2)^2/3*(n/3)) = e^(-n/36) = 0.001}}}

===> {{{-n/36 = ln0.001}}}  ===> {{{n =-36*ln0.001  = 248.68}}}, or {{{highlight(n = 249)}}}, rounded to the nearest whole number.