A fair coin is tossed n times. Let X be the n | vedclass

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(D) The probability of getting $k$ heads in $n$ tosses of a fair coin is given by the binomial distribution: $P(X = k) = \binom{n}{k} (\frac{1}{2})^n$

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None of these

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(D) The probability of getting $k$ heads in $n$ tosses of a fair coin is given by the binomial distribution: $P(X = k) = \binom{n}{k} (\frac{1}{2})^n$.Given that $P(X = 4), P(X = 5), P(X = 6)$ are in $H.P.$,their reciprocals must be in $A.P.$Thus,$\frac{1}{P(X = 4)}, \frac{1}{P(X = 5)}, \frac{1}{P(X = 6)}$ are in $A.P.$This implies $\frac{2}{P(X = 5)} = \frac{1}{P(X = 4)} + \frac{1}{P(X = 6)}$.Substituting the values: $\frac{2}{\binom{n}{5}} = \frac{1}{\binom{n}{4}} + \frac{1}{\binom{n}{6}}$.Using $\binom{n}{k} = \frac{n!}{k!(n-k)!}$,we get $\frac{2 	imes 5!(n-5)!}{n!} = \frac{4!(n-4)!}{n!} + \frac{6!(n-6)!}{n!}$.Multiplying by $\frac{n!}{4!(n-6)!}$,we obtain: $2 	imes 5 	imes (n-5) = (n-4)(n-5) + 6 	imes 5$.$10n - 50 = n^2 - 9n + 20 + 30$.$n^2 - 19n + 100 = 0$.The discriminant $D = (-19)^2 - 4(1)(100) = 361 - 400 = -39$.Since $D < 0$,there is no real integer solution for $n$. Thus,the correct option is $D$.

$A$ fair coin is tossed $n$ times. Let $X$ be the number of times head is observed. If $P(X = 4), P(X = 5)$ and $P(X = 6)$ are in $H.P.$,then $n$ is equal to

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