<p>A point with coordinates (x, y) is chosen at random from the unit square. What is the probability that the point satisfies \(y^2 \leq x\)?</p>
Step-by-Step Solution
Key Concept: The region where y² ≤ x within the unit square [0,1]×[0,1] is bounded by the parabola x = y². Calculate the area under this curve and divide by the total unit square area (1).
<p><strong>Step 1:</strong> Identify the region. The condition y² ≤ x describes the area to the right of the parabola x = y² within the unit square [0,1]×[0,1].</p><p><strong>Step 2:</strong> Set up the integral. The parabola x = y² intersects the unit square from y = 0 to y = 1 (since when y = 1, x = 1). For a fixed y ∈ [0,1], x ranges from y² to 1.</p><p><strong>Step 3:</strong> Calculate the area of the favorable region:</p><p>$$A = \int_0^1 (1 - y^2) \, dy = \left[y - \frac{y^3}{3}\right]_0^1 = 1 - \frac{1}{3} = \frac{2}{3}$$</p><p><strong>Step 4:</strong> Find the probability. Since the unit square has area 1, the probability is:</p><p>$$P = \frac{2/3}{1} = \frac{2}{3} ≈ 0.667$$</p><p>∴ Answer: <strong>0.67</strong></p>
Correct Answer: 0.67