Question

One hundred identical coins each with probability $p$ of showing up heads are tossed once. If

$0<p<1$ and the probability of heads showing on $50$ coins is equal to that of heads showing on $51$ coins, then find the value of $p$.

Answer 1

We have ${}^{100}{C}_{50}{p}^{50}{(1-p)}^{50}{=}^{100}{C}_{51}{p}^{51}{(1-p)}^{49}$

or $\frac{1-p}{p}=\frac{100!}{51!49!}\times \frac{50!50!}{100!}=\frac{50}{51}$

$\Rightarrow 51-51p=50p$

$\Rightarrow 51=50p+51p$

$\Rightarrow 101p=51$

$\therefore p=\frac{51}{101}$