Probability
Region of probability conditions
Grade None
Question:
<p>Let <em>P(x)</em> denote the probability of the occurrence of event <em>x</em>. Plot all those points <em>(x, y) = (P(A), P(B))</em> in a plane which satisfy the conditions, <em>P(A ∪ B) ≥ 3/4</em> and <em>1/8 ≤ P(A ∩ B) ≤ 3/8</em>. The shaded region representing the solution is bounded by which of the following?</p>
<p>\(x + y = \frac{7}{8}\) and \(x + y = \frac{11}{8}\) with \(0 \le x \le 1,\ 0 \le y \le 1\)</p>
<p>\(x + y \ge \frac{7}{8}\) and \(x + y \le \frac{11}{8}\) with \(0 \le x \le 1,\ 0 \le y \le 1\)</p>
<p>\(x + y = \frac{3}{4}\) and \(x + y = \frac{3}{8}\) with \(0 \le x \le 1,\ 0 \le y \le 1\)</p>
<p>None of the above</p>
Step-by-Step Solution
Key Concept: Use P(A ∪ B) = P(A) + P(B) - P(A ∩ B) to express the constraint as a linear inequality in x and y, then find the region bounded by this line and the constraints on P(A ∩ B) which acts as a parameter.
<p><strong>Step 1:</strong> Use the addition rule: P(A ∪ B) = P(A) + P(B) - P(A ∩ B), or x + y - P(A ∩ B) ≥ 3/4</p><p><strong>Step 2:</strong> Rearrange to get: y ≥ 3/4 - x + P(A ∩ B)</p><p><strong>Step 3:</strong> Since 1/8 ≤ P(A ∩ B) ≤ 3/8, the constraint becomes a region between two parallel lines:</p><ul><li>Lower boundary: y = 7/8 - x (when P(A ∩ B) = 1/8)</li><li>Upper boundary: y = 9/8 - x (when P(A ∩ B) = 3/8)</li></ul><p><strong>Step 4:</strong> Additional constraints: 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, and P(A ∩ B) ≤ min(x,y)</p><p><strong>Step 5:</strong> The shaded region is bounded by the lines y = 7/8 - x and y = 9/8 - x, along with the constraints x + y ≤ 1 (from probability bounds) and the hyperbolic/curved boundaries from P(A ∩ B) ≤ min(x,y).</p><p>∴ Answer: A</p>
Correct Answer: A