<p>A bag contains 3 red and 3 green balls and a person draws out 3 at random. He then drops 3 blue balls into the bag and again draws out 3 at random. The chance that the 3 later balls being all of different colors is</p>
Step-by-Step Solution
Key Concept: The composition of balls in the second draw depends on what was drawn in the first draw. After removing 3 balls and adding 3 blue balls, we have 6 balls total (3 remaining from original + 3 blue). We need to use conditional probability: P(all different colors in 2nd draw) = Σ P(all different | 1st draw outcome) × P(1st draw outcome).
<p><strong>Step 1: Identify possible first draws</strong></p><p>First draw of 3 balls from 3R, 3G:</p><ul><li>3R, 0G: P₁ = C(3,3)×C(3,0)/C(6,3) = 1/20</li><li>2R, 1G: P₂ = C(3,2)×C(3,1)/C(6,3) = 9/20</li><li>1R, 2G: P₃ = C(3,1)×C(3,2)/C(6,3) = 9/20</li><li>0R, 3G: P₄ = C(3,0)×C(3,3)/C(6,3) = 1/20</li></ul><p><strong>Step 2: After first draw, add 3 blue balls</strong></p><p>After each case, bag has 6 balls total (3 original remaining + 3 blue):</p><ul><li>Case 1 (3R,0G drawn): Bag has 0R, 3G, 3B → P(1R,1G,1B) = 0</li><li>Case 2 (2R,1G drawn): Bag has 1R, 2G, 3B → P(1R,1G,1B) = C(1,1)×C(2,1)×C(3,1)/C(6,3) = 6/20</li><li>Case 3 (1R,2G drawn): Bag has 2R, 1G, 3B → P(1R,1G,1B) = C(2,1)×C(1,1)×C(3,1)/C(6,3) = 6/20</li><li>Case 4 (0R,3G drawn): Bag has 3R, 0G, 3B → P(1R,1G,1B) = 0</li></ul><p><strong>Step 3: Apply total probability</strong></p><p>P(all different in 2nd draw) = (1/20)×0 + (9/20)×(6/20) + (9/20)×(6/20) + (1/20)×0</p><p>= (9/20)×(6/20) + (9/20)×(6/20) = 2×(54/400) = 108/400 = 27/100</p><p>∴ Answer: C</p>
Correct Answer: C