Probability
Conditional Probability
Grade 12
Question:
<p>Box I contains 5 red and 2 blue balls, while box II contains 2 red and 6 blue balls. A fair coin is tossed. If it turns up head, a ball is drawn from box I, else a ball is drawn from box II. The probability that the ball drawn is from box I, if it is blue, is</p>
<p>(a) \(\frac{27}{56}\)</p>
<p>(b) \(\frac{8}{29}\)</p>
<p>(c) \(\frac{21}{29}\)</p>
<p>(d) \(\frac{29}{56}\)</p>
Step-by-Step Solution
Key Concept: Apply Bayes' theorem to find the posterior probability that the ball came from Box I given that it is blue.
<p><strong>Step 1:</strong> Use Bayes' theorem. P(Box I | Blue) = P(Blue | Box I) × P(Box I) / P(Blue)</p><p><strong>Step 2:</strong> P(Blue | Box I) = 2/7, P(Box I) = 1/2</p><p><strong>Step 3:</strong> P(Blue | Box II) = 6/8 = 3/4, P(Box II) = 1/2</p><p><strong>Step 4:</strong> P(Blue) = (2/7)(1/2) + (3/4)(1/2) = 1/7 + 3/8 = 29/56</p><p><strong>Step 5:</strong> P(Box I | Blue) = (2/7 × 1/2) / (29/56) = (1/7) / (29/56) = 8/29</p>
Correct Answer: B