Q. One hundred identical coins, each with probability $p$ of showing heads are tossed once. If $0 < p < 1$ and the probability of head showing on 50 coins is equal to that of head showing on 51 coins, the value of $p$ is
A) $\frac{1}{2}$
B) $\frac{51}{101}$
C) $\frac{49}{101}$
D) None of these
Explanation
(b)
Let $X$ be binomial variate with parameter $n = 100\ and\ P$
Since, $P(X = 50) = P(X = 51)$ [given$\rbrack$
$\Rightarrow {\ ^{100}C}_{50}p^{50}(1 - p)^{50} = {\ ^{100}C}_{51}p^{51}(1 - p)^{49}$
$\Rightarrow \ \frac{100!}{50!50!} \times \frac{51!49!}{100!} = \frac{p}{1 - p}$
$\Rightarrow \ \frac{51}{50} = \frac{p}{1 - p}$
$\Rightarrow \ p = \frac{51}{101}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.