Question

$201$ coins each with probability $P(0<P<1)$ of showing head are tossed together. If the probability of getting $100$ heads is equal to the probability of getting $101$ heads, then the value of $P$ is

• A. $\frac{1}{4}$
• B. $\frac{1}{3}$
• C. $\frac{1}{2}$
• D. $\frac{1}{6}$

Answer 1

The correct option is C $\frac{1}{2}$

Probability of $100$ heads and $101$ tails $=$ probability of $101$ heads and $100$ tails (out of $201$ coins)
Thus,
${}^{201}{C}_{101}\ast P^{101}\ast {(1-P)}^{100}{=}^{201}{C}_{100}\ast P^{100}\ast {(1-P)}^{101}=>P=1-P=>P=\frac{1}{2}$