Probability Question 148
Question: If X has binomial distribution with mean np and variance npq, then $ \frac{P(X=k)}{P(X=k-1)} $ is
[Pb. CET 2004]
Options:
A) $ \frac{n-k}{k-1}.\frac{p}{q} $
B) $ \frac{n-k+1}{k}.\frac{p}{q} $
C) $ \frac{n+1}{k}.\frac{q}{p} $
D) $ \frac{n-1}{k+1}.\frac{q}{p} $
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Answer:
Correct Answer: B
Solution:
Here mean = np and variance = npq
$ \therefore $ $ \frac{P(X=k)}{P(X=k-1)}=\frac{^{n}C_{k}{{(p)}^{k}}{{(q)}^{n-k}}}{^{n}{C_{k-1}}{{(p)}^{k-1}}{{(q)}^{n-k+1}}}=\frac{^{n}C_{k}}{^{n}{C_{k-1}}}.\frac{p}{q} $
$ \therefore $ $ \frac{P(X=k)}{P(X=k-1)}=\frac{n-k+1}{k}.\frac{p}{q} $ .
BETA
