We need to find
\[
P(X_1 + X_2 + X_3 < \tfrac{1}{2}),
\]
where each \(X_i\) is independent and uniformly distributed on \([0,1]\)
This means we must integrate 1 over the region satisfying
\[
x_1 + x_2 + x_3 < \tfrac{1}{2}, \quad 0 \le x_1, x_2, x_3 \le 1
\]
We will integrate in the order \(x_3 \rightarrow x_2 \rightarrow x_1\)
Limits for \(x_1\)
Since all \(x_i \ge 0\), \(x_1 \ge 0\)
Just having \(x_1 > \tfrac{1}{2}\) will make the sum \(x_1 + x_2 + x_3 > \tfrac{1}{2}\)
Hence the boundary of $x_1$ is,
\[
0 \le x_1 \le \tfrac{1}{2}
\]
Limits for \(x_2\)
Fix \(x_1\). The inequality becomes
\[x_2 + x_3 < \tfrac{1}{2} - x_1\]
Just having \(x_2 \ge \frac{1}{2}-x_1\) will make the sum $x_2 + x_3 > \tfrac{1}{2} - x_1$
Hence the boundary of $x_2$ is,
\[0 \le x_2 \le \tfrac{1}{2} - x_1 \]
Limits for \(x_3\)
Fix \(x_1, x_2\). The inequality gives
\[x_3 < \tfrac{1}{2} - x_1 - x_2\]
Just having \(x_3 \ge \frac{1}{2}-x_1-x_2\) will make the sum $x_3 > \tfrac{1}{2} - x_1 - x_2$
Hence the boundary of $x_3$ is,
\[
0 \le x_3 \le \tfrac{1}{2} - x_1 - x_2
\]
Combining all limits, the required probability is
\[
P(X_1 + X_2 + X_3 < \tfrac{1}{2})
= \int_{x_1 = 0}^{1/2} \int_{x_2 = 0}^{1/2 - x_1} \int_{x_3 = 0}^{1/2 - x_1 - x_2} dx_3\, dx_2\, dx_1
\]
Evaluating the innermost integral:
\[
\int_0^{1/2 - x_1 - x_2} dx_3 = \tfrac{1}{2} - x_1 - x_2
\]
Now integrate with respect to \(x_2\):
\[
\int_0^{1/2 - x_1} (\tfrac{1}{2} - x_1 - x_2)\,dx_2
= \tfrac{1}{2} (\tfrac{1}{2} - x_1)^2
\]
Finally, integrate with respect to \(x_1\):
\[
\tfrac{1}{2} \int_0^{1/2} (\tfrac{1}{2} - x_1)^2\,dx_1
= \tfrac{1}{2} \left[\frac{(\tfrac{1}{2} - x_1)^3}{3}\right]_0^{1/2}
= \frac{1}{48}
\]
