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MHT CET · Maths · Probability

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

  1. A \(\frac{1}{2}\)
  2. B \(\frac{49}{101}\)
  3. C \(\frac{50}{101}\)
  4. D \(\frac{51}{101}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{51}{101}\)

Step-by-step Solution

Detailed explanation

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