Q. One hundred identical coins, each with probability $p$ or 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
A) $\frac{1}{2}$
B) $\frac{49}{101}$
C) $\frac{50}{101}$
D) $\frac{51}{101}$
Explanation
(d)
Let $X$ be the number of coins showing heads. Let $X$ be a binomial variate with parameters $n = 100$ and $p$ probability of success. According to question
$P(X = 50) = P(X = 51)$
$\Rightarrow {\ ^{100}C}_{50}{\ p}^{50}(1 - p)^{50} = {\ ^{100}C}_{51}(p)^{51}(1 - p)^{49}$
$\Rightarrow \ \frac{(100)!}{(50!)(50!)} \bullet \frac{(51!) \times (49!)}{100!} = \frac{p}{1 - p}$
$\Rightarrow \ \frac{p}{1 - p} = \frac{51}{50}\ \Rightarrow p = \frac{51}{101}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.