<p>The probability of drawing one white ball and one green ball from the first urn is \(\frac{1}{5}\). The probability of drawing one white ball and one green ball from the second urn is \(\frac{1}{3}\). The probability of drawing one white ball and one green ball from the third urn is \(\frac{2}{11}\). Therefore, the probability that the third urn was chosen is \(\frac{a}{b}\). Find \(a + b\).</p>
Step-by-Step Solution
Key Concept: Use Bayes' theorem: P(third urn | event) = P(event | third urn) × P(third urn) / P(event). The total probability P(event) is the sum over all urns, assuming equal prior probabilities (1/3 each) for selecting any urn.
<p><strong>Step 1:</strong> Identify the conditional probabilities given:<br>P(W∩G | Urn 1) = 1/5<br>P(W∩G | Urn 2) = 1/3<br>P(W∩G | Urn 3) = 2/11<br>Assume equal prior: P(Urn i) = 1/3 for each urn.</p><p><strong>Step 2:</strong> Apply the law of total probability:<br>P(W∩G) = P(W∩G|Urn 1)·P(Urn 1) + P(W∩G|Urn 2)·P(Urn 2) + P(W∩G|Urn 3)·P(Urn 3)<br>P(W∩G) = (1/5)(1/3) + (1/3)(1/3) + (2/11)(1/3)<br>P(W∩G) = 1/15 + 1/9 + 2/33</p><p><strong>Step 3:</strong> Find common denominator (LCM of 15, 9, 33 is 495):<br>P(W∩G) = 33/495 + 55/495 + 30/495 = 118/495</p><p><strong>Step 4:</strong> Apply Bayes' theorem:<br>P(Urn 3 | W∩G) = P(W∩G|Urn 3)·P(Urn 3) / P(W∩G)<br>P(Urn 3 | W∩G) = (2/11 × 1/3) / (118/495)<br>P(Urn 3 | W∩G) = (2/33) / (118/495)<br>P(Urn 3 | W∩G) = (2/33) × (495/118)<br>P(Urn 3 | W∩G) = (2 × 495)/(33 × 118) = 990/3894 = 15/59</p><p><strong>Step 5:</strong> Verify the fraction is in lowest form: gcd(15, 59) = 1 ✓<br>Therefore a = 15, b = 59<br>∴ Answer: a + b = 15 + 59 = <strong>74</strong></p>
Correct Answer: 74