If X has a binomial distribution with mean np | vedclass

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(B) The probability mass function for a binomial distribution is given by $P(X = k) = ^nC_k p^k q^{n-k}$.We need to evaluate the ratio $\frac{P(X = k)

Vedclass · Accepted Answer

$\frac{n - k + 1}{k} \cdot \frac{p}{q}$

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(B) The probability mass function for a binomial distribution is given by $P(X = k) = ^nC_k p^k q^{n-k}$.We need to evaluate the ratio $\frac{P(X = k)}{P(X = k - 1)}$.$\frac{P(X = k)}{P(X = k - 1)} = \frac{^nC_k p^k q^{n-k}}{^nC_{k-1} p^{k-1} q^{n-k+1}}$Using the formula for combinations,$^nC_k = \frac{n!}{k!(n-k)!}$,we have:$\frac{^nC_k}{^nC_{k-1}} = \frac{n!}{k!(n-k)!} \cdot \frac{(k-1)!(n-k+1)!}{n!} = \frac{n-k+1}{k}$Substituting this back into the ratio:$\frac{P(X = k)}{P(X = k - 1)} = \left( \frac{n-k+1}{k} \right) \cdot \frac{p^k q^{n-k}}{p^{k-1} q^{n-k+1}} = \frac{n-k+1}{k} \cdot \frac{p}{q}$.

If $X$ has a binomial distribution with mean $np$ and variance $npq$,then $\frac{P(X = k)}{P(X = k - 1)}$ is

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