Mathematics - Probability Question with Solution | TestHub

$\begin{array}{rl}\text{ Mean }& =np=4\\ \text{ Variance }& =npq=3\\ \Rightarrow q& =\frac{3}{4}\end{array}$ and $p=1-q=1-\frac{3}{4}=\frac{1}{4}$ $\Rightarrow n=16$ Now, Mode of Binomial distribution is given by $(n+1)P$ Case IIf $(n+1)P=\mathrm{Integer}(I)$, then Mode $=\{I,I-1\}$ Case II If $(n+1)P\ne \mathrm{Integer}(I+f)$, then Mode $=\{I\}$ $\begin{array}{rl}\therefore (n+1)P& =(16+1)\frac{1}{4}\\ & =\frac{17}{4}=4.25\\ & =4+0.25\\ \Rightarrow \text{ Mode }& =4\end{array}$