<p>If <em>A</em> and <em>B</em> are two events such that \(P(A \cup B) = P(A \cap B)\), then the incorrect statement amongst the following statements is:</p>
<p><em>A</em> and <em>B</em> are equally likely</p>
<p>\(P(A \cap B') = 0\)</p>
<p>\(P(A' \cup B) = 0\)</p>
<p>\(P(A) + P(B) = 1\)</p>
Step-by-Step Solution
Key Concept: When P(A ∪ B) = P(A ∩ B), the events must satisfy P(A) + P(B) = P(A ∩ B), which severely constrains their relationship. This equality holds only when A and B are either identical or one is empty, making most standard probability relationships fail.
<p><strong>Step 1:</strong> Use the formula P(A ∪ B) = P(A) + P(B) - P(A ∩ B)</p><p><strong>Step 2:</strong> Given that P(A ∪ B) = P(A ∩ B), substitute:</p><p>P(A ∩ B) = P(A) + P(B) - P(A ∩ B)</p><p><strong>Step 3:</strong> Rearrange to get:</p><p>2P(A ∩ B) = P(A) + P(B)</p><p><strong>Step 4:</strong> This means P(A) + P(B) - 2P(A ∩ B) = 0, or equivalently [P(A) - P(A ∩ B)] + [P(B) - P(A ∩ B)] = 0</p><p><strong>Step 5:</strong> Since probabilities are non-negative, this forces P(A) = P(A ∩ B) and P(B) = P(A ∩ B), meaning A ⊆ B and B ⊆ A, so A = B</p><p><strong>Step 6:</strong> Therefore A and B must be identical events. Any statement implying they are distinct, independent (unless trivial), or mutually exclusive will be incorrect.</p><p>∴ Answer: D</p>
Correct Answer: D