In a binomial distribution, the mean is $$4$$ and the variance is $$3$$. What is the mode ?

Solution :
As given, $$np = 4$$  and $$npq = 3$$
[where $$p$$ is the probability of success and $$q$$ is the probability of failure for an event to occur, and $$'n'$$ is the number of trials]
$$\eqalign{ & \Rightarrow q = \frac{{npq}}{{np}} = \frac{3}{4} \cr & {\text{Also, }}p = 1 - q = 1 - \frac{3}{4} = \frac{1}{4} \cr & \therefore \,n = 16 \cr} $$
In a binomial distribution, the value of $$r$$ for which $$P\left( {X = r} \right)$$   is maximum is the mode of binomial distribution.
$$\eqalign{ & {\text{Hence, }}\left( {n + 1} \right)p - 1 \leqslant r \leqslant \left( {n + 1} \right)p \cr & \Rightarrow \frac{{17}}{4} - 1 \leqslant r \leqslant \frac{{17}}{4} \cr & \Rightarrow \frac{{13}}{4} \leqslant r \leqslant \frac{{17}}{4} \cr & \Rightarrow 3.25 \leqslant r \leqslant 4.25 \cr & \Rightarrow r = 4 \cr} $$