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(A) Given that $X$ follows a binomial distribution,the probability mass function is $P(X=r) = { }^n C_r p^r q^{n-r}$,where $q = 1-p$.The ratio is give

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$\frac{1}{2}$

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(A) Given that $X$ follows a binomial distribution,the probability mass function is $P(X=r) = { }^n C_r p^r q^{n-r}$,where $q = 1-p$.The ratio is given by:$\frac{P(X=r)}{P(X=n-r)} = \frac{{ }^n C_r p^r q^{n-r}}{{ }^n C_{n-r} p^{n-r} q^r}$Since ${ }^n C_r = { }^n C_{n-r}$,the expression simplifies to:$\frac{P(X=r)}{P(X=n-r)} = \frac{p^r q^{n-r}}{p^{n-r} q^r} = \left(\frac{p}{q}\right)^r \left(\frac{q}{p}\right)^{n-r} = \left(\frac{q}{p}\right)^{n-2r}$For this expression to be independent of $n$,the base of the exponent involving $n$ must be $1$.Thus,$\frac{q}{p} = 1$,which implies $q = p$.Since $p + q = 1$,we have $p + p = 1$,which gives $2p = 1$.Therefore,$p = \frac{1}{2}$.

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 $n$ for every $r$,then $p$ is equal to

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