An urn contains $X$ white balls and $120-X$ red balls while $X$ is a random variable satisfying $\mathbb{E}X=30$. We take one ball from this urn with equal probabilities.
What is the probability that this ball is white?
I have a solution that says that $\frac{1}{4}$ is the answer since this is the expected value of the random variable $\frac{X}{120}$ but I don't understand why is that true.
Please any help.
On
It is a variation on the Law of Total Probability.
$\begin{split}\mathsf P(Y{=}y\mid X{=}x) &= x/120&&{ \textsf{Where $Y$ is the count for white balls drawn.}\\\textsf{and $X$ the count for white balls in the urn of 120}}\\[2ex]\mathsf P(Y{=}y)&=\mathsf E(\mathbf 1_{Y=y})&\quad&\textsf{definition}\\&=\mathsf E(\mathsf E(\mathbf 1_{Y=y}\mid X))&&\textsf{law of total expectation}\\ &= \mathsf E(\mathsf P(Y{=}y\mid X))&&\textsf{definition}\\ &= \mathsf E(X/120) &&\textsf{evaluation} \\ &= 30/120 &&\textsf{given }\mathsf E(X)=30\\ &= 1/4\end{split}$
Let $p_i=P(X=i)$. Then we are told that $$E[X]=\sum_{i=0}^{120}ip_i=30$$
If there are exactly $i$ white balls the probability of drawing one is $\frac i{120}$. Thus the total probability of drawing a white ball is $$\sum_{i=0}^{120}\frac i{120}p_i=\frac 1{120}\times \sum_{i=0}^{120}ip_i=\frac {30}{120}=\frac 14$$ as desired.