<p>If \(0 < P(A) \leq 1\), \(0 < P(B) < 1\) then \(P(\overline{A/B})\) is equal to</p>
<p>(a) \(\frac{1 - P(A \cup B)}{P(B)}\)</p>
<p>(b) \(1 - P(A/B)\)</p>
<p>(c) \(\frac{1 - P(A \cup B)}{P(B)}\)</p>
<p>(d) \(\frac{P(A)}{P(B)}\)</p>
Step-by-Step Solution
Key Concept: Use the property that for continuous uniform distributions, P(X ≤ a) = a when X is uniformly distributed on [0,1]. The intersection of two independent uniform random variables follows a specific pattern that requires careful analysis of the region where both conditions are satisfied simultaneously.
<p><strong>Step 1:</strong> Recognize that X and Y are uniformly distributed on (0,p), so they form a joint uniform distribution on the square (0,p) × (0,p) with total area p².</p><p><strong>Step 2:</strong> The condition X ≤ Y is satisfied in the region above the line y = x within the square [0,p]². This region (a triangle) has vertices at (0,0), (0,p), and (p,p).</p><p><strong>Step 3:</strong> Calculate the area of the triangle where X ≤ Y: Area = ½ × p × p = p²/2</p><p><strong>Step 4:</strong> Apply the geometric probability formula: P(X ≤ Y) = (Area where X ≤ Y)/(Total area) = (p²/2)/p² = 1/2</p><p><strong>Step 5:</strong> Notice this result is independent of p, which is the key insight—the probability remains constant regardless of the upper bound.</p><p>∴ Answer: A</p>
Correct Answer: A