expectation of  product of  independent random variable it is a problem from my exam praperation sheet 
Let $U$ , $Y$ be independent random variables. Here $U$ is uniformly distributed on $(0 , 1)$ .
whereas $Y$ is $\frac{1}{4} \delta(0) + \frac{3}{4} \delta (1)$ .  Let $X := UY$ .
(a)Find $E[X]$
(b)Graph $F$ and find $P \left( 0 \le X \le \frac{2}{3} \right)$
can any one tell me how to solve this problem? Thanks
 A: For $E(X)$, one can note that since $Y$ and $U$ are independent, $E(X)=E(UY)=E(U)E(Y)$.
A fancier derivation which can be useful when we are dealing with something more complicated than sum or product is to use conditional expectation. We have $E(X|Y=0)=0$, and $E(X|Y=1)=E(U)=1/2$. Thus by what is sometimes called the Law of Total Expectation,
$E(X)=(0)(1/4)+(1/2)(3/4)$.
For the second part, the event $X\le 2/3$ can happen in two ways: (i) $Y=0$ and $X\le 2/3$, or (ii) $Y=1$ and $X\le 2/3$. 
(i) The conditional probability that $X\le 2/3$, given that $Y=0$, is $1$. (ii) The conditional probability that  $X\le 2/3$, given that $Y=1$, is $2/3$. So our required probability is
$$\frac{1}{4}\cdot 1 +\frac{3}{4}\cdot\frac{2}{3}.$$
Remark: It may be useful to tell a story about what is going on. We are thirsty, and end up playing the following game. An amount of water, uniformly distributed in $[0,1]$ (litres) is put into a jar.  We then flip a coin that has probability $1/4$ of landing head, and $3/4$ of landing tail. If it lands head, we don't get the water in the jar. If it lands tail, we do get the water. 
Then the amount of water we get has the same distribution as the random variable $UY$ of the question. So in the first part, we are calculating the mean amount of water we get, and in the second part we are calculating the probability that the amount of water we get is $\le \frac{2}{3}$ (of a litre).    
