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(D) Let $X$ be the number of heads in $n = 100$ tosses. $X$ follows a binomial distribution $B(n, p)$.Given that $P(X = 50) = P(X = 51)$.The probabili

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$\frac{51}{101}$

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(D) Let $X$ be the number of heads in $n = 100$ tosses. $X$ follows a binomial distribution $B(n, p)$.Given that $P(X = 50) = P(X = 51)$.The probability mass function is $P(X = k) = {}^{n}C_{k} p^k (1-p)^{n-k}$.So,${}^{100}C_{50} p^{50} (1-p)^{50} = {}^{100}C_{51} p^{51} (1-p)^{49}$.Dividing both sides by $p^{50} (1-p)^{49}$,we get:${}^{100}C_{50} (1-p) = {}^{100}C_{51} p$.$\frac{1-p}{p} = \frac{{}^{100}C_{51}}{{}^{100}C_{50}} = \frac{100!}{51! 49!} 	imes \frac{50! 50!}{100!} = \frac{50}{51}$.$51(1-p) = 50p$.$51 - 51p = 50p$.$101p = 51$.$p = \frac{51}{101}$.

One hundred identical coins,each with probability $p$ of showing 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

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