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

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$\frac{1}{2}$

$\frac{49}{101}$

$\frac{50}{101}$

$\frac{51}{101}$

answer is D.

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Detailed Solution

Let $X$ be the number of coins showing heads. Then $X ~ B (100, p) .$ We have

$P (X = 51) = P (X = 50)$
$\Rightarrow^{100} C_{51} p^{51} q^{49} =^{100} C_{50} p^{50} q^{50}$ (where $q = 1 - p$ )

$\begin{array}{l} \Rightarrow \frac{p}{q} = \frac{100!}{50! 50!} \frac{51! 49!}{100!} \Rightarrow \frac{p}{1 - p} = \frac{51}{50} \\ \Rightarrow 50 p = 51 - 51 p \Rightarrow 101 p = 51 \\ \Rightarrow p = 51 / 101 . \end{array}$

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