• # question_answer One hundred identical coins, each with probability p, of showing up heads are tossed. If 0 < p < 1 and the probability of heads showing on 50 coins is equal to that of the heads showing on 51 coins, then p = A) $\frac{1}{2}$                                    B) $\frac{49}{101}$ C) $\frac{50}{101}$                                             D) $\frac{51}{101}$

Here $n=100,p=p,q=1-p$ Given, P(50) = P(5 $\Rightarrow$$^{100}{{C}_{50}}{{p}^{50}}{{(1-p)}^{50}}{{=}^{10}}{{C}_{51}}{{p}^{51}}{{(1-p)}^{49}}$ $\Rightarrow$$\frac{100!}{50!50!}(1-p)=\frac{100!}{51!49!}p$$\Rightarrow$$51(1-p)=50p$ $\Rightarrow$$p=\frac{51}{101}$