Question

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

• A. $\frac{n-k}{k-1}·\frac{p}{q}$
• B. $\frac{n-k+1}{k}·\frac{p}{q}$
• C. $\frac{n+1}{k}·\frac{p}{q}$
• D. $\frac{n-1}{k+1}·\frac{p}{q}$

Answer 1

The correct option is B

$\frac{n-k+1}{k}·\frac{p}{q}$

Explanation for the correct option.

Find the value of the expression.

For binomial distribution for parameters $n$ and $p$, $P\left(X=r\right)$ is expanded as: $P\left(X=r\right)={}^{n}C_r{p}^r{q}^{n-r}$ where $q=1-p$.

So, the value of the expression $\frac{P\left(X=k\right)}{P\left(X=k-1\right)}$ is given as:

$\begin{array}{rcl}\frac{P\left(X=k\right)}{P\left(X=k-1\right)}& =& \frac{{}^{n}C_k{p}^k{q}^{n-k}}{{}^{n}C_{k-1}{p}^{k-1}{q}^{n-k+1}}\\ & =& \frac{{\displaystyle \frac{n!}{k!\left(n-k\right)!}}}{{\displaystyle \frac{n!}{(k-1)!(n-k+1)!}}}{p}^{k-k+1}{q}^{n-k-n+k-1}\\ & =& \frac{(k-1)!(n-k+1)!}{k!\left(n-k\right)!}p{q}^{-1}\\ & =& \frac{(k-1)!(n-k+1)(n-k)!}{k(k-1)!(n-k)!}\frac{p}{q}\\ & =& \frac{n-k+1}{k-1}·\frac{p}{q}\end{array}$

Thus the value of $\frac{P\left(X=k\right)}{P\left(X=k-1\right)}$ is $\frac{n-k+1}{k}·\frac{p}{q}$.

Hence, the correct option is B.