<p>Each of the \(n\) urns contains 4 white and 6 black balls. The \((n+1)\)th urn contains 5 white and 5 black balls. One of the \(n+1\) urns is chosen at random and two balls are drawn from it without replacement. Both the balls turn out to be black. If the probability that the \((n+1)\)th urn was chosen to draw the balls is \(1/16\), then find the value of \(n\).</p>
Step-by-Step Solution
Key Concept: Use Bayes' theorem: the probability of choosing urn (n+1) given two black balls are drawn equals P(2 black | urn n+1) × P(urn n+1) divided by the total probability of drawing 2 black balls from any urn. Set this equal to 1/16 and solve for n.
<p><strong>Step 1:</strong> Calculate P(2 black balls | urn k) for k = 1 to n and for urn (n+1).</p><p>For each of the first n urns (4 white, 6 black): P(2 black) = C(6,2)/C(10,2) = 15/45 = 1/3</p><p>For urn (n+1) (5 white, 5 black): P(2 black) = C(5,2)/C(10,2) = 10/45 = 2/9</p><p><strong>Step 2:</strong> Apply Bayes' theorem. Each urn chosen with probability 1/(n+1).</p><p>P(urn n+1 | 2 black) = [P(2 black | urn n+1) × P(urn n+1)] / [Total probability of 2 black]</p><p><strong>Step 3:</strong> Calculate total probability of drawing 2 black balls:</p><p>P(2 black) = n urns × (1/(n+1)) × (1/3) + 1 urn × (1/(n+1)) × (2/9)</p><p>P(2 black) = (1/(n+1)) × [n/3 + 2/9] = (1/(n+1)) × [(3n + 2)/9]</p><p><strong>Step 4:</strong> Apply Bayes' formula:</p><p>P(urn n+1 | 2 black) = [(2/9) × (1/(n+1))] / [(1/(n+1)) × (3n+2)/9]</p><p>= (2/9) / [(3n+2)/9] = 2/(3n+2)</p><p><strong>Step 5:</strong> Set equal to given probability:</p><p>2/(3n+2) = 1/16</p><p>32 = 3n + 2</p><p>3n = 30</p><p>∴ Answer: n = 10</p>
Correct Answer: 10