Question

One hundred identical coins each with a probability $p$ of showing up heads are tossed once. If $0<p<1$ and the probability of heads showing on $50$ coins are equal to that of heads showing on $51$ coins, then 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

$\frac{51}{101}$

Explanation for the correct option

Let $X$ be the number of coins that show heads

Then $X~B(100,p)$

On the basis of the given information

$$\begin{gathered} P(X=50)=P(X=51) \\ \Rightarrow {}^{100}C_{50}{p}^{50}{q}^{50}={}^{100}C_{51}{p}^{51}{q}^{49}\left[\because {}^{n}C_r{p}^r{q}^{n-r}\right] \end{gathered}$$

On solving, we get:

$$\begin{gathered} \Rightarrow \frac{51}{50}=\frac{p}{q} \\ \Rightarrow \frac{51}{50}=\frac{p}{1-p}[\because p+q=1] \\ \Rightarrow p=\frac{51}{101} \end{gathered}$$

Hence, the correct answer is option (D).