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 the value of p is

$\text{ Let }p\text{ be the probability of one coin showing head. Then the probability of one coin showing tail is }1-p\text{. According to }$

$\text{ question, the coin is tossed }100\text{ times and probability of }50\text{ coins showing head is equal to the probability of }51\text{ coins }$$\text{ showing head. }$

$\text{ Using binomial probability distribution }P(X=r){=}^n{C}_r{p}^r{q}^{n-r}\text{, }$

we get,

$\begin{array}{l}{ }^{100}{\mathrm{C}}_5{\mathrm{p}}^{50}(1-\mathrm{p}{)}^{50}{=}^{100}{\mathrm{C}}_{51}{\mathrm{p}}^{51}(1-\mathrm{p}{)}^{49}\\ \mathrm{or} \frac{1-\mathrm{p}}{\mathrm{p}}=\frac{100{\mathrm{c}}_{51}}{100{\mathrm{c}}_{50}}=\frac{50!50!}{51!49!}=\frac{50}{51}\\ \mathrm{or} 51-51\mathrm{p}=50\mathrm{p}\\ \mathrm{or} 101\mathrm{p}=51\text{ or }\mathrm{p}=\frac{51}{101}\end{array}$