Q.If $X\sim B(n, p)$ such that $4P(X=4)=P(X=2)$ and $n=6$, find the distribution, mean and Standard Deviation of X.

Uses the given probability relation to find $p$, then reads off the full binomial distribution and its mean/SD formulas.

1. For $X\sim B(n,p)$ with $n=6$: $P(X=k)=\binom6kp^kq^{6-k}$, where $q=1-p$.
2. Given $4P(X=4)=P(X=2)$: $4\binom64p^4q^2=\binom62p^2q^4$.
3. $\binom64=\binom62=15$ (symmetry of binomial coefficients), so $4(15)p^4q^2=15p^2q^4\Rightarrow4p^4q^2=p^2q^4$.
4. Divide both sides by $p^2q^2$ (both nonzero, since $0<p<1$): $4p^2=q^2$.
5. Taking positive square roots (since $p,q>0$): $2p=q$.
6. Using $p+q=1$: $p+2p=1\Rightarrow3p=1\Rightarrow p=\dfrac13$, so $q=\dfrac23$.
7. Distribution: $P(X=k)=\binom6k\left(\dfrac13\right)^k\left(\dfrac23\right)^{6-k}=\dfrac{\binom6k\,2^{6-k}}{729}$ for $k=0,1,\dots,6$: $P(0)=\dfrac{64}{729},\ P(1)=\dfrac{192}{729},\ P(2)=\dfrac{240}{729},\ P(3)=\dfrac{160}{729},\ P(4)=\dfrac{60}{729},\ P(5)=\dfrac{12}{729},\ P(6)=\dfrac{1}{729}$. …