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Q.

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 20 coins is equal to that of showing on 21 coins then the value of p is

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a

$\frac{1}{2}$

b

$\frac{21}{101}$

c

$\frac{50}{101}$

d

$\frac{51}{101}$

answer is B.

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Detailed Solution

$p (X = 20) = p (x = 21)$

$\Rightarrow^{100} C_{20} q^{80} p^{20} =^{100} C_{21} q^{79} p^{21} \Rightarrow q = \frac{80}{21} p \Rightarrow p = \frac{21}{101}$

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