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

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$\frac{49}{101}$

$\frac{50}{101}$

$\frac{51}{101}$

$\frac{1}{2}$

answer is D.

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Detailed Solution

$100_{C_{_{50}}} q^{50} p^{50} = 100_{C_{_{51}}} q^{49} P^{51}$

$\Rightarrow q = \frac{100_{C_{_{51}}}}{100_{C_{_{50}}}} p$

$\Rightarrow q = \frac{50}{51} p$

$\Rightarrow 51 (1 - p) = 50$

$∴ p = \frac{51}{101}$

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