Suppose X follows a binomial distribution wit | vedclass

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(A) The probability mass function for a binomial distribution is given by $P(X=r) = { }^n C_r p^r q^{n-r}$, where $q = 1-p$. We are given the ratio $\

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$\frac{1}{2}$

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(A) The probability mass function for a binomial distribution is given by $P(X=r) = { }^n C_r p^r q^{n-r}$, where $q = 1-p$. We are given the ratio $\frac{P(X=r)}{P(X=n-r)}$. Substituting the formula: $\frac{P(X=r)}{P(X=n-r)} = \frac{{ }^n C_r p^r q^{n-r}}{{ }^n C_{n-r} p^{n-r} q^r}$. Since ${ }^n C_r = { }^n C_{n-r}$, the expression simplifies to: $\frac{P(X=r)}{P(X=n-r)} = \frac{p^r q^{n-r}}{p^{n-r} q^r} = \left(\frac{p}{q}\right)^{r - (n-r)} = \left(\frac{p}{q}\right)^{2r-n} = \left(\frac{q}{p}\right)^{n-2r}$. For this expression to be independent of $n$, the base must be $1$ (or the exponent must be $0$, but it must hold for all $r$). 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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