Solution: Recall that for a random variable $X$ with the probability density function (PDF) $f(x)$, the expect4d value is given by \[ E [X] = \int\limits_{-\infty }^{\infty } xf(x) \mathrm{d} x . \] Here at first we need to determine the PDF for the distribution $X = \frac{X_1}{X_1 + X_2 + X_3}$ and then use the above integral to find the expectation. Instead doing this, we can use the following observation. Note that since $X_1,X_2,X_3 \sim U(0,1),$ the radome variables $\frac{X_1}{X_1 + X_2 + X_3}, \frac{X_2}{X_1 + X_2 + X_3} $ and $\frac{X_3}{X_1 + X_2 + X_3}$ have the same distribution. Therefore, \[ E \left[ \frac{X_1}{X_1 + X_2 + X_3} \right] = E \left[ \frac{X_2}{X_1 + X_2 + X_3} \right] = E \left[ \frac{X_3}{X_1 + X_2 + X_3} \right]. \] Let us denote $E \left[ \frac{X_i}{X_1 + X_2 + X_3} \right] = \mu_i = \mu $. Then, \begin{align*} & \mu _1 + \mu _2 + \mu _3 = E \left[ \frac{X_1 + X_2 + X_3}{X_1 + X_2 + X_3} \right] = 1 \\ \implies & 3\mu = 1 \implies \mu = \frac{1}{3}. \end{align*} Thus, \[ E \left[ \frac{X_1}{X_1 + X_2 + X_3} \right] = \frac{1}{3}. \]