expected value and variance of a binomial variable

Question

If $X \sim \operatorname{Bin}(10, \theta)$, $p(2)=0.1$ and $p(3)=0.2$, find $\operatorname{E}(X)$ and $\operatorname{V}(X)$

I'm a little lost here so the first thing I did was:

$$p(2)={10 \choose 2}\theta^2(1-\theta)^8=0.1$$ then, $$45\theta^2(1-\theta)^8=0.1 \rightarrow \theta^2(1-\theta)^8=\frac{1}{450}$$ and,

$$\theta(1-\theta)^4=\sqrt{\frac{1}{450}}$$ I did the same with $p(3)=0.2$, but I'm not very sure that's what I am supposed to do.

Answer 1

Guide:$$\theta^2 (1-\theta)^8 = a$$ and $$\theta^3(1-\theta)^7 = b$$

implies $$\frac{1-\theta}{\theta}=\frac{a}{b}$$

which can be converted into a linear equation in $\theta$.

After you solve for for $\theta$, you should be able to solve the problem.

Answer 2

\begin{equation} 2=\frac{0.2}{0.1}=\frac{p(3)}{p(2)} = \frac{\binom{10}{3} \theta^3 (1-\theta)^7}{\binom{10}{2} \theta^2 (1-\theta)^8} = \frac{8\theta}{3(1-\theta)}, \end{equation} then solve for $\theta$.