Probability
Total Probability Theorem
Grade 12
Question:
<p>A = even that the item came from lot \(A\); \(P(A) = \dfrac{3}{7}\). \(B\) = item came from \(B\); \(P(B) = \dfrac{4}{7}\). \(D\) = item from mixed lot 'C' is defective. \(P(D) = P(D \cap A) + P(D \cap B) = P(A) \cdot P(D/A) + P(B) \cdot P(D/A)\). Find \(P(D)\).</p>
<p>\(\dfrac{29}{56}\)</p>
<p>\(\dfrac{3}{7}\)</p>
<p>\(\dfrac{29}{56}\)</p>
<p>\(\dfrac{4}{7}\)</p>
Step-by-Step Solution
Key Concept: Use the law of total probability by partitioning the sample space into mutually exclusive events (lots A and B), then sum the conditional probabilities weighted by their prior probabilities.
<p><strong>Step 1:</strong> Identify the partition: The item comes from either lot A or lot B, which are mutually exclusive and exhaustive events.</p><p><strong>Step 2:</strong> Apply the law of total probability: Since D (defective item) can occur with items from A or B:</p><p>P(D) = P(D ∩ A) + P(D ∩ B)</p><p><strong>Step 3:</strong> Express using conditional probability:</p><p>P(D) = P(A) · P(D|A) + P(B) · P(D|B)</p><p><strong>Step 4:</strong> Substitute known values (note: the problem states P(D|A) should be given; assuming standard values or the problem requires the formula structure):</p><p>P(D) = (3/7) · P(D|A) + (4/7) · P(D|B)</p><p><strong>Step 5:</strong> If specific conditional probabilities are provided in the complete problem, substitute and calculate the numerical answer.</p><p>∴ The framework P(D) = P(A) · P(D|A) + P(B) · P(D|B) is the essential result using total probability.</p>
Correct Answer: C