A fair coin is tossed a fixed number of times | vedclass

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(C) Let $n$ be the number of tosses. For a fair coin, the probability of heads $p = \frac{1}{2}$ and the probability of tails $q = \frac{1}{2}$.Using

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$\frac{21}{128}$

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(C) Let $n$ be the number of tosses. For a fair coin, the probability of heads $p = \frac{1}{2}$ and the probability of tails $q = \frac{1}{2}$.Using the binomial distribution formula $P(X=k) = {}^nC_k p^k q^{n-k}$, we have:$P(X=5) = P(X=4)$${}^nC_5 (\frac{1}{2})^5 (\frac{1}{2})^{n-5} = {}^nC_4 (\frac{1}{2})^4 (\frac{1}{2})^{n-4}$${}^nC_5 (\frac{1}{2})^n = {}^nC_4 (\frac{1}{2})^n$${}^nC_5 = {}^nC_4$Using the property ${}^nC_r = {}^nC_{n-r}$, we get $n = 5 + 4 = 9$.Now, we need to find the probability of getting $6$ heads:$P(X=6) = {}^9C_6 (\frac{1}{2})^6 (\frac{1}{2})^3 = {}^9C_3 (\frac{1}{2})^9$$P(X=6) = \frac{9 	imes 8 	imes 7}{3 	imes 2 	imes 1} 	imes \frac{1}{512} = 84 	imes \frac{1}{512} = \frac{21}{128}$.

$A$ fair coin is tossed a fixed number of times. If the probability of getting $5$ heads is equal to the probability of getting $4$ heads, then the probability of getting $6$ heads is

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