Number of toss to reject it is a fair coin You toss the coin n times, and you have observed 60% of times, it is head.
How large n needs to be in order to achieve 95% confidence that it is not a fair coin?
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Attempt: Basically use Bionmial distribution, and but I have no idea how to take account of the 60% number into my calculation.
 A: The question implies that you are to use the number of heads tossed as the test statistic for a null-hypothesis significance test of the hypothesis $\ H_0\ $ that the coin being tossed is fair, and find the smallest value of $\ n\ $ for which $\ 0.6n\ $ lies in a critical region of significance $5\%$ (or, equivalently, outside the complementary confidence interval of $95\%$).
Unless you have a strong reason to believe that the coin could only be biassed towards heads (which is not suggested by anything in the wording of the question), you should be using a two-tailed test, for which the confidence interval would have the form $\ \left[\lfloor0.4n\rfloor+1, \lceil0.6n\rceil-1\right]\ $.  So if $\ N_h\ $ is the number of heads tossed, you have to find the smallest value of $\ n\ $ such that
\begin{align}
P\left(N_h\in \left[\lfloor0.4n\rfloor+1, \lceil0.6n\rceil-1\right]\big|H_0\right)&=\frac{1}{2^n}\sum_{k=\lfloor0.4n\rfloor+1}^{\lceil0.6n\rceil-1}{n\choose k}\\
&\ge0.95
\end{align}
Under the null hypthesis, the mean of $\ N_h\ $ is $\ \frac{n}{2}\ $, and its variance is $\ \frac{n}{4}\ $, so $\ 0.6n\ $ is $\ 0.2\sqrt{n}\ $ standard deviations above the mean, and for large enough $\ n\ $ the probability that $\ N_h\ $ lies in the given interval can be well approximated by $\ \mathcal{N}(0,1)\big(0.2\sqrt{n}\big)-$$\mathcal{N}\big(0,1\big)(-0.2\sqrt{n})\ $, or, equivalently, by $\ 2\mathcal{N}(0,1)\big(0.2\sqrt{n}\big)-1\ $. So if $\ u\ $ is the (unique) positive number such that $\ \mathcal{N}(0,1)(u)=0.975\ $, then $\ n=\left(\frac{u}{0.2}\right)^2\ $ will be a reasonable approximation for the value you require.
