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(B) Given that a fair coin is tossed $K$ times,the probability of getting $X$ heads follows a binomial distribution with $p = \frac{1}{2}$ and $q = \f

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$5$

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(B) Given that a fair coin is tossed $K$ times,the probability of getting $X$ heads follows a binomial distribution with $p = \frac{1}{2}$ and $q = \frac{1}{2}$.Given $P(X=4) = P(X=6)$.Using the binomial probability formula $P(X=r) = {}^K C_r p^r q^{K-r}$,we have:${}^K C_4 (\frac{1}{2})^K = {}^K C_6 (\frac{1}{2})^K$${}^K C_4 = {}^K C_6$Since ${}^n C_x = {}^n C_y$ implies $n = x + y$ (for $x 
eq y$),we get $K = 4 + 6 = 10$.For a binomial distribution with $p = q = \frac{1}{2}$,the probability $P(X=r)$ is maximum at the mean value.For $n = 10$,the maximum probability occurs at $r = \frac{n}{2} = \frac{10}{2} = 5$.

$A$ fair coin is tossed $K$ times such that the probability of getting $4$ heads is equal to that of getting $6$ heads. If the probability is maximum for getting $r$ heads,then $r=$

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