Question

One hundred identical coins, each with probability $p$ of showing 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, the value of $p$ is

• A. $1/2$
• B. $51/101$
• C. $49/101$
• D. $3/101$

Answer 1

Answer: (B)

$51/101$

Let $X$ be the number of coins showing heads. Then $X$ follows a binomial distribution with parameters $n=100$ and $p$.
Since $P(X=50)=P(X=51)$,

we get, ${}^{100}{C}_{50}{p}^{50}(1-p{)}^{50}{=}^{100}{C}_{51}{l}^{51}(1-p^{49}$
$\Rightarrow \frac{100!}{50!50!}\cdot \frac{51!49!}{100!}=\frac{p}{1-p}\Rightarrow \frac{51}{50}=\frac{p}{1-p}$
$\Rightarrow 51-51p=50p\Rightarrow p=51/101$.