Question

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 heads showing on $51$ coins, the value of $p$ is

• A. $\frac{1}{2}$
• B. $\frac{49}{101}$
• C. $\frac{50}{101}$
• D. $\frac{51}{101}$

Answer 1

The correct option is D $\frac{51}{101}$

Let $X\sim B(100,p)$ be the number of coins showing heads and let $q=1-p$.
Then, since $P(X=51)=P(X=50)$,
we have ${}^{100}{C}_{51}\left(p^{51}\right)\left(q^{49}\right){=}^{100}{C}_{50}\left(p^{50}\right)\left(q^{50}\right)$
$\Rightarrow \frac{p}{q}=\left(\frac{100!}{50!50!}\right)\left(\frac{51!49!}{100!}\right)$
$\Rightarrow \frac{p}{1-p}=\frac{51}{50}\Rightarrow 50p=51-51p\Rightarrow p=\frac{51}{101}$