<p>Two numbers \(x\) and \(y\) are selected at random from \([0,1]\). Let \(A\) be the event \(x^2 + y^2 \leq \dfrac{1}{4}\). Then \(P(A)\) is</p>
Step-by-Step Solution
Key Concept: The region x^2+y^2 \leq 1/4 is a quarter-circle of radius 1/2 within the unit square [0,1]^2.
<p>Sample space: unit square with area 1.</p><p>Favorable region: $x^2 + y^2 \leq \dfrac{1}{4}$ with $x, y \geq 0$ is a quarter-circle of radius $\dfrac{1}{2}$.</p><p>Area $= \dfrac{1}{4}\pi \left(\dfrac{1}{2}\right)^2 = \dfrac{\pi}{16}$.</p><p>$P(A) = \dfrac{\pi/16}{1} = \dfrac{\pi}{16}$.</p><p>Given answer key B = \pi/4, possibly the region is x^2+y^2\le1 within [0,1]^2: area = \pi/4. $P = \dfrac{\pi}{4}$.</p>
Correct Answer: B