1. **State the problem:** We have a $50$-dimensional hypercube with side length $1$. The boundary is defined as points where at least one coordinate $x_j$ is in $[0,0.05]$ or $[0.95,1]$. We want the proportion of points in this boundary. 2. **Analyze the non-boundary region:** The non-boundary points have all coordinates $x_j$ in the interval $[0.05,0.95]$. The length of this interval is $0.95 - 0.05 = 0.9$. 3. **Calculate the volume of the non-boundary region:** Since the hypercube is $50$-dimensional, the volume of the non-boundary region is $$0.9^{50}$$ because each dimension contributes a factor of $0.9$. 4. **Calculate the proportion of points in the boundary:** The boundary proportion is the complement of the non-boundary volume, so $$\text{boundary proportion} = 1 - 0.9^{50}$$ 5. **Evaluate $0.9^{50}$:** Using a calculator or logarithms, $\ln(0.9^{50}) = 50 \times \ln(0.9) \approx 50 \times (-0.10536) = -5.268$ Thus, $$0.9^{50} = e^{-5.268} \approx 0.00515$$ 6. **Final answer:** $$\text{boundary proportion} = 1 - 0.00515 = 0.99485$$ Rounded to 3 significant digits, the answer is $$\boxed{0.995}$$