Question

For a binomial distribution if $P=\frac{2}{3}$ and $n=10$ the probability of mode is

• A. ${}^{10}{C}_4{\left(\frac{1}{3}\right)}^7{\left(\frac{2}{3}\right)}^3$
• B. ${}^{10}{C}_7{\left(\frac{2}{3}\right)}^7{\left(\frac{1}{3}\right)}^3$
• C. ${\left(\frac{2}{3}\right)}^7{\left(\frac{1}{3}\right)}^3$
• D. ${\left(\frac{1}{3}\right)}^7{\left(\frac{2}{3}\right)}^3$

Answer 1

The correct option is B ${}^{10}{C}_7{\left(\frac{2}{3}\right)}^7{\left(\frac{1}{3}\right)}^3$

$(n+1)\left(p\right)$
$=(10+1)\left(\frac{2}{3}\right)$
$=\frac{22}{3}$
Hence $(n+1)\left(p\right)$ is non-integral.
Therefore
mode$=\left[\frac{22}{3}\right]$; where $\left[x\right]$ denotes greatest integer function.
mode$=\left[7.33\right]$
$=7$
Hence the require value is
${}^{10}{C}_7\left(\frac{2}{3}{)}^7\right(\frac{1}{3}{)}^3$