<p>In a bag there are six balls of unknown colours, three balls are drawn at random and found to be all black. Find the probability that no black ball is left in the bag. (Or, Find the probability that the bag contained exactly 3 black balls.)</p>
Step-by-Step Solution
Key Concept: Use Bayes' theorem with the given evidence (3 black balls drawn). The posterior probability depends on the likelihood of drawing 3 black balls from each possible composition of the bag, weighted by prior probabilities.
<p><strong>Step 1: Set up possible scenarios</strong></p><p>The bag contains 6 balls. Let B_k = event that exactly k balls are black (k = 0,1,2,3,4,5,6).</p><p>Prior probability: P(B_k) = 1/7 for each k (uniform prior over 7 possibilities).</p><p><strong>Step 2: Find likelihood of drawing 3 black balls</strong></p><p>Event E: "3 balls drawn are all black"</p><p>P(E|B_k) = C(k,3)·C(6-k,0) / C(6,3) if k ≥ 3, else 0</p><p>• P(E|B_0) = 0</p><p>• P(E|B_1) = 0</p><p>• P(E|B_2) = 0</p><p>• P(E|B_3) = C(3,3)·C(3,0)/C(6,3) = 1·1/20 = 1/20</p><p>• P(E|B_4) = C(4,3)·C(2,0)/C(6,3) = 4·1/20 = 4/20</p><p>• P(E|B_5) = C(5,3)·C(1,0)/C(6,3) = 10·1/20 = 10/20</p><p>• P(E|B_6) = C(6,3)·C(0,0)/C(6,3) = 20·1/20 = 20/20</p><p><strong>Step 3: Apply law of total probability</strong></p><p>P(E) = Σ P(E|B_k)·P(B_k) = (1/7)·(1/20 + 4/20 + 10/20 + 20/20) = (1/7)·(35/20) = 35/140 = 1/4</p><p><strong>Step 4: Apply Bayes' theorem</strong></p><p>P(B_3|E) = P(E|B_3)·P(B_3) / P(E) = (1/20)·(1/7) / (1/4) = (1/140)·(4/1) = 4/140 = 1/35</p><p>∴ Answer: C (Probability = 1/35)</p>
Correct Answer: C