Question

For a binomial distribution if $P=\frac{1}{4},n=20$, the probability of mode is

• A. ${}^{20}{C}_5{\left(\frac{3}{4}\right)}^5$
• B. ${}^{20}{C}_5{\left(\frac{1}{4}\right)}^5{\left(\frac{3}{4}\right)}^{15}$
• C. ${}^{10}{C}_{10}{\left(\frac{1}{4}\right)}^{10}{\left(\frac{3}{4}\right)}^{10}$
• D. ${}^{10}{C}_{10}{\left(\frac{3}{4}\right)}^{10}$

Answer 1

The correct option is B ${}^{20}{C}_5{\left(\frac{1}{4}\right)}^5{\left(\frac{3}{4}\right)}^{15}$

Given:

$p=\frac{1}{4},n=20$

For binomial distribution, mode $=(n+1)\left(p\right)$

$=(20+1)\cdot \frac{1}{4}$ $=\frac{21}{4}$
Hence,

$(n+1)\left(p\right)$ is a non-integral value.

Therefore,
mode $=\left[\frac{21}{4}\right]$, where $[\cdot ]$ denotes greatest integer function.
mode $=\left[5.25\right]=5$
Hence, the required value of probability of mode is
${}^{20}{C}_5{\left(\frac{1}{4}\right)}^5{\left(\frac{3}{4}\right)}^{15}$

Hence, option A.