<p><em>A</em> and <em>B</em> are two candidates seeking admission to IIT. The probability that <em>A</em> is selected is 0.5 and the probability that <em>A</em> and <em>B</em> are selected is at most 0.3. Is it possible that the probability of <em>B</em> getting selected is 0.9?</p>
Step-by-Step Solution
Key Concept: We need to check if P(B) = 0.9 is consistent with the given constraints using the property that P(A ∩ B) ≤ P(A) and the upper bound P(A ∩ B) ≤ 0.3.
<p><strong>Step 1:</strong> Identify the given information.</p><p>• P(A) = 0.5</p><p>• P(A ∩ B) ≤ 0.3</p><p>• We need to check if P(B) = 0.9 is possible.</p><p><strong>Step 2:</strong> Apply the fundamental probability constraint.</p><p>By the definition of conditional probability and basic probability theory:</p><p>P(A ∩ B) ≤ P(A) = 0.5</p><p>P(A ∩ B) ≤ P(B)</p><p><strong>Step 3:</strong> Use the given upper bound constraint.</p><p>We are given that P(A ∩ B) ≤ 0.3, which is a stricter constraint than P(A ∩ B) ≤ P(A).</p><p><strong>Step 4:</strong> Check consistency with P(B) = 0.9.</p><p>If P(B) = 0.9 and P(A ∩ B) ≤ 0.3, then:</p><p>P(A ∩ B) ≤ 0.3 ... (given)</p><p>P(A ∩ B) ≤ P(B) = 0.9 ... (always true)</p><p><strong>Step 5:</strong> Apply the principle P(A ∩ B) = P(A) · P(B|A).</p><p>We have P(A ∩ B) ≤ 0.3 and P(A) = 0.5.</p><p>This means: P(B|A) ≤ 0.3/0.5 = 0.6</p><p>So the conditional probability of B given A is at most 0.6.</p><p><strong>Step 6:</strong> Use the law of total probability.</p><p>P(B) = P(B|A)·P(A) + P(B|A^c)·P(A^c)</p><p>0.9 = P(B|A)·(0.5) + P(B|A^c)·(0.5)</p><p>1.8 = 0.5·P(B|A) + 0.5·P(B|A^c)</p><p><strong>Step 7:</strong> Check if this is achievable.</p><p>Since P(B|A) ≤ 0.6, and P(B|A^c) ≤ 1, we have:</p><p>P(B) ≤ 0.6(0.5) + 1(0.5) = 0.3 + 0.5 = 0.8</p><p>But we need P(B) = 0.9, which violates this upper bound.</p><p><strong>Step 8:</strong> Conclusion.</p><p>It is NOT possible for P(B) = 0.9 given the constraints P(A) = 0.5 and P(A ∩ B) ≤ 0.3.</p><p>∴ Answer: B</p>
Correct Answer: B