The probability of a coin showing head is p a | vedclass

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(C) Let $X$ be the number of heads in $100$ tosses. $X$ follows a binomial distribution $B(n, p)$ where $n = 100$. The probability of getting $k$ head

Vedclass · Accepted Answer

$\frac{51}{101}$

Vedclass · Answer

(C) Let $X$ be the number of heads in $100$ tosses. $X$ follows a binomial distribution $B(n, p)$ where $n = 100$. The probability of getting $k$ heads is given by $P(X=k) = {}^{n}C_{k} p^{k} (1-p)^{n-k}$. Given that $P(X=50) = P(X=51)$, we have: ${}^{100}C_{50} p^{50} (1-p)^{50} = {}^{100}C_{51} p^{51} (1-p)^{49}$ Dividing both sides by $p^{50} (1-p)^{49}$, we get: ${}^{100}C_{50} (1-p) = {}^{100}C_{51} p$ Using the formula ${}^{n}C_{r} = \frac{n!}{r!(n-r)!}$, we have: $\frac{100!}{50! 50!} (1-p) = \frac{100!}{51! 49!} p$ $\frac{1-p}{50!} = \frac{p}{51 	imes 50! 	imes \frac{49!}{50!}} \Rightarrow \frac{1-p}{50} = \frac{p}{51}$ $51(1-p) = 50p$ $51 - 51p = 50p$ $101p = 51$ $p = \frac{51}{101}$

The probability of a coin showing head is $p$ and then $100$ such coins are tossed. If the probability of $50$ coins showing head is same as the probability of $51$ coins showing head, then $p$ equals

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