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Question: If X has binomial distribution with mean np and variance npq, then ![](https://cdn.pureessence.tech/...

If X has binomial distribution with mean np and variance npq, then is

A

n−kk−1⋅pq\frac { \mathrm { n } - \mathrm { k } } { \mathrm { k } - 1 } \cdot \frac { \mathrm { p } } { \mathrm { q } }

B

C

n+1k⋅qp\frac { \mathrm { n } + 1 } { \mathrm { k } } \cdot \frac { \mathrm { q } } { \mathrm { p } }

D

n−1k+1⋅qp\frac { \mathrm { n } - 1 } { \mathrm { k } + 1 } \cdot \frac { \mathrm { q } } { \mathrm { p } }

Answer

Explanation

Solution

Here mean = np and variance= npq

∴ P(X=k)P(X=k−1)=nCk(p)k(q)n−knCk−1(p)k−1(q)n−k+1=nCknCk−1⋅pq\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 } } \cdot \frac { p } { q }

∴ P(X=k)P(X=k−1)=n−k+1k⋅pq\frac { \mathrm { P } ( \mathrm { X } = \mathrm { k } ) } { \mathrm { P } ( \mathrm { X } = \mathrm { k } - 1 ) } = \frac { \mathrm { n } - \mathrm { k } + 1 } { \mathrm { k } } \cdot \frac { \mathrm { p } } { \mathrm { q } }