<p>Let AB be a line segment of length <em>a</em> divided at P and Q such that AP = x and BQ = y. A point is selected at random inside the triangle formed by the conditions. What is the probability that x < a/2, y < a/2, PQ < a/2? (Given answer: 0.25)</p>
Step-by-Step Solution
Key Concept: The sample space is defined by the constraints AP = x, BQ = y on segment AB of length a, which means PQ = a - x - y. We need to find the probability that three simultaneous conditions (x < a/2, y < a/2, PQ < a/2) are satisfied, which translates to finding the area of intersection of three half-planes in the (x,y) coordinate system divided by total feasible area.
<p><strong>Step 1: Identify the sample space</strong></p><p>Since AP = x and BQ = y on segment AB of length a, we have:</p><ul><li>0 ≤ x ≤ a (point P is on AB)</li><li>0 ≤ y ≤ a (point Q is on AB)</li><li>PQ = a - x - y ≥ 0, so x + y ≤ a (P and Q don't overlap)</li></ul><p>Sample space is a triangle with vertices (0,0), (a,0), (0,a) with area = a²/2</p></p><p><strong>Step 2: Apply the three conditions simultaneously</strong></p><ul><li>Condition 1: x < a/2</li><li>Condition 2: y < a/2</li><li>Condition 3: PQ < a/2 ⟹ a - x - y < a/2 ⟹ x + y > a/2</li></ul></p><p><strong>Step 3: Find the favorable region</strong></p><p>The favorable region is bounded by:</p><ul><li>x = a/2 (vertical line)</li><li>y = a/2 (horizontal line)</li><li>x + y = a/2 (diagonal line)</li><li>x + y = a (original constraint)</li></ul><p>The intersection forms a region where x < a/2, y < a/2, and a/2 < x + y < a.</p><p>This region is a triangle with vertices: (0, a/2), (a/2, 0), and (a/2, a/2)</p><p>Area of favorable region = (1/2) × (a/2) × (a/2) = a²/8</p></p><p><strong>Step 4: Calculate probability</strong></p><p>Probability = (a²/8)/(a²/2) = (a²/8) × (2/a²) = 1/4 = <strong>0.25</strong></p>
Correct Answer: 0.25