Suppose X follows a binomial distribution wit | vedclass

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(A) The probability mass function of a binomial distribution is given by $P(X = k) = {}^nC_k p^k (1 - p)^{n-k}$.We are given the ratio $\frac{P(X = r)

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$p = \frac{1}{2}$

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(A) The probability mass function of a binomial distribution is given by $P(X = k) = {}^nC_k p^k (1 - p)^{n-k}$.We are given the ratio $\frac{P(X = r)}{P(X = n - r)}$.Substituting the formula,we get:$\frac{P(X = r)}{P(X = n - r)} = \frac{{}^nC_r p^r (1 - p)^{n-r}}{{}^nC_{n-r} p^{n-r} (1 - p)^r}$.Since ${}^nC_r = {}^nC_{n-r}$,the binomial coefficients cancel out:$\frac{P(X = r)}{P(X = n - r)} = \frac{p^r (1 - p)^{n-r}}{p^{n-r} (1 - p)^r} = \frac{p^r}{(1 - p)^r} \cdot \frac{(1 - p)^{n-r}}{p^{n-r}} = \left( \frac{p}{1 - p} ight)^r \cdot \left( \frac{1 - p}{p} ight)^{n-r} = \left( \frac{p}{1 - p} ight)^r \cdot \left( \frac{p}{1 - p} ight)^{-(n-r)} = \left( \frac{p}{1 - p} ight)^{r - n + r} = \left( \frac{p}{1 - p} ight)^{2r - n}$.Alternatively,$\frac{P(X = r)}{P(X = n - r)} = \left( \frac{1 - p}{p} ight)^{n - 2r}$.For this expression to be independent of $n$ and $r$,the base must be $1$ (since $1^{	ext{anything}} = 1$).Thus,$\frac{1 - p}{p} = 1 \implies 1 - p = p \implies 2p = 1 \implies p = \frac{1}{2}$.

Suppose $X$ follows a binomial distribution with parameters $n$ and $p$,where $0 < p < 1.$ If $\frac{P(X = r)}{P(X = n - r)}$ is independent of $n$ and $r$,then

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