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(D) Let $X$ be the number of heads,which follows a binomial distribution $B(n, p)$ with $n = 100$. The probability of getting $k$ heads is given by $P

Vedclass · Accepted Answer

$\frac{51}{101}$

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(D) Let $X$ be the number of heads,which follows a binomial distribution $B(n, p)$ with $n = 100$. The probability of getting $k$ heads is given by $P(X=k) = {}^{n}C_{k} p^{k} (1-p)^{n-k}$. We are given that $P(X=50) = P(X=51)$. Substituting the values,we get: ${}^{100}C_{50} p^{50} (1-p)^{50} = {}^{100}C_{51} p^{51} (1-p)^{49}$. Dividing both sides by $p^{50} (1-p)^{49}$,we get: ${}^{100}C_{50} (1-p) = {}^{100}C_{51} p$. Rearranging to solve for $p$: $\frac{1-p}{p} = \frac{{}^{100}C_{51}}{{}^{100}C_{50}} = \frac{100!}{51! 49!} 	imes \frac{50! 50!}{100!} = \frac{50}{51}$. Thus,$51(1-p) = 50p$. $51 - 51p = 50p$. $101p = 51$. $p = \frac{51}{101}$.

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

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