One hundred identical coins, each with probability p , of showing up heads are tossed once. If $0 \lt \mathrm{p} \lt 1$ 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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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { We have }{ }^{100} \mathrm{C}_{50} \mathrm{p}^{50}(1-\mathrm{p})^{50}={ }^{100} \mathrm{C}_{51} \mathrm{p}^{51}(1-\mathrm{p})^{49} \ & \Rightarrow \frac{1-\mathrm{p}}{\mathrm{p}}=\frac{100!}{51!49!} 	imes \frac{50!.50!}{100!}=\frac{50}{51} \ & \Rightarrow 51-51 \mathrm{p}=50 \mathrm{p} \ & \Rightarrow \mathrm{p}=\frac{51}{101}\end{aligned}$

One hundred identical coins, each with probability p , of showing up heads are tossed once. If $0 \lt…

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