If X sim B(n, p) then frac{P(X=k)}{P(X=k-1)}= | vedclass

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(D) For a binomial distribution $X \sim B(n, p)$,the probability mass function is given by $P(X=k) = \binom{n}{k} p^k q^{n-k}$,where $q = 1-p$. We nee

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

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(D) For a binomial distribution $X \sim B(n, p)$,the probability mass function is given by $P(X=k) = \binom{n}{k} p^k q^{n-k}$,where $q = 1-p$. We need to find the ratio $\frac{P(X=k)}{P(X=k-1)}$. $P(X=k) = \frac{n!}{k!(n-k)!} p^k q^{n-k}$ $P(X=k-1) = \frac{n!}{(k-1)!(n-k+1)!} p^{k-1} q^{n-k+1}$ Taking the ratio: $\frac{P(X=k)}{P(X=k-1)} = \frac{n!}{k!(n-k)!} p^k q^{n-k} \cdot \frac{(k-1)!(n-k+1)!}{n! p^{k-1} q^{n-k+1}}$ $= \frac{(n-k+1)!}{(n-k)!} \cdot \frac{(k-1)!}{k!} \cdot \frac{p^k}{p^{k-1}} \cdot \frac{q^{n-k}}{q^{n-k+1}}$ $= (n-k+1) \cdot \frac{1}{k} \cdot p \cdot \frac{1}{q}$ $= \frac{n-k+1}{k} \cdot \frac{p}{q}$ Thus,the correct option is $D$.

If $X \sim B(n, p)$ then $\frac{P(X=k)}{P(X=k-1)}=$

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