Question

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

• A. $\frac{1}{2}$
• B. $\frac{51}{101}$
• C. $\frac{49}{101}$
• D. None of these

Answer 1

The correct option is B

$\frac{51}{101}$

Explanation for the correct option:

Given. Total coin is hundred.

The probability of the head showing on $50$ coins are equal to that of the head shown on $51$ coins

Applying binomial distribution for $100$ trials using the formula ${}^{n}C_r{\left(p\right)}^r{\left(q\right)}^{n-r}$ we get

$$\begin{gathered} \Rightarrow {}^{100}{C}_{50}{p}^{50}{(1-p)}^{50}{=}^{100}{C}_{51}{p}^{51}{(1-p)}^{49} \\ \Rightarrow \frac{(1-p)}{50}=\frac{p}{51} \\ \Rightarrow \frac{\left(1-p\right)}{50}=\frac{p}{51} \\ \Rightarrow 51-51p=50p \\ \Rightarrow 101p=51 \\ \Rightarrow p=\frac{51}{101} \end{gathered}$$

Hence, the correct answer is option (B).