Probability
Bayes' Theorem
Grade 12
Question:
<p>There are \((n+1)\) urns, each containing some balls. Let \(E_1\) denote the event that one of the first \(n\) urns is chosen and \(E_2\) denote the event that the \((n+1)\)th urn is selected. \(A\) denotes the event that two balls drawn are black. Then \(p(E_1) = \dfrac{n}{n+1}\), \(P(E_2) = \dfrac{1}{n+1}\), \(P(A/E_1) = \dfrac{^6C_2}{^{10}C_2} = \dfrac{1}{3}\) and \(P(A/E_2) = \dfrac{^5C_2}{^{10}C_2} = \dfrac{2}{9}\). Using Bayes' Theorem, if \(P(E_2/A) = \dfrac{1}{16}\), find the value of \(n\).</p>
Step-by-Step Solution
Key Concept: Apply Bayes' Theorem: P(E₂/A) = P(A/E₂)·P(E₂) / P(A), where P(A) is found using the law of total probability P(A) = P(A/E₁)·P(E₁) + P(A/E₂)·P(E₂). The given conditional probabilities and prior probabilities must be combined correctly.
<p><strong>Step 1:</strong> Find P(A) using the law of total probability:</p><p>P(A) = P(A/E₁)·P(E₁) + P(A/E₂)·P(E₂)</p><p>P(A) = (1/3)·(n/(n+1)) + (2/9)·(1/(n+1))</p><p>P(A) = [1/(n+1)]·[(n/3) + (2/9)]</p><p>P(A) = [1/(n+1)]·[(3n + 2)/9] = (3n + 2)/(9(n+1))</p><p><strong>Step 2:</strong> Apply Bayes' Theorem:</p><p>P(E₂/A) = P(A/E₂)·P(E₂) / P(A)</p><p>1/16 = [(2/9)·(1/(n+1))] / [(3n + 2)/(9(n+1))]</p><p><strong>Step 3:</strong> Simplify:</p><p>1/16 = [(2/9)·(1/(n+1))] · [9(n+1)/(3n + 2)]</p><p>1/16 = 2/(3n + 2)</p><p><strong>Step 4:</strong> Cross-multiply and solve for n:</p><p>3n + 2 = 32</p><p>3n = 30</p><p>n = 10</p><p>∴ Answer: <strong>10</strong></p>
Correct Answer: 10