Can I calculate $E(X^2)$ if I know that $X$~$B(1, \frac{1}{3})$ (binomial distribution)

Question

If I know that I have $X$~$B(1, \frac{1}{3})$ (binomial distribution),

Can I calculate (the expectancy) $E(X^2)$ ?

(That's a part of a question, that if I can calculate it, it will make the question faster).

An explanation would be appreciated.

Answer 1

$$ EX^2 = 0^2 \times \frac{2}{3} + 1^2 \times \frac{1}{3} = \frac{1}{3}. $$ Generally, you can observe that in your case $EX^k = \frac{1}{3}$, for all $k > 0$, as $$ EX^k = 0^k \times\frac{2}{3} + 1^k \times \frac{1}{3} = \frac{1}{3}. $$

Answer 2

Note that if $X$~$B(n,p)$ then $E(X)=np$ and $V(X)=npq$ where $q=1-p$. And $$E(X^2)=E(X)^2+V(X) \tag{1}$$

You can calculate the desired value from $(1)$.

Answer 3

Note that $X$ only takes values in $\{0,1\}$ so that $X^2=X$.