<p>In a bag there are 6 balls of which 3 are white and 3 are black. They are drawn successively with replacement. What is the chance that the colours are alternate?</p>
Step-by-Step Solution
Key Concept: With replacement, each draw is independent with P(white) = P(black) = 1/2. For alternating colors in n draws, multiply the probability of each individual outcome following the alternating pattern.
<p><strong>Step 1:</strong> Identify the setup. We have 6 balls (3 white, 3 black), drawing with replacement. P(white) = 3/6 = 1/2, P(black) = 3/6 = 1/2.</p><p><strong>Step 2:</strong> For alternating colors in 4 draws, there are two patterns:</p><p>• Pattern 1: WBWB → P = (1/2)(1/2)(1/2)(1/2) = 1/16</p><p>• Pattern 2: BWBW → P = (1/2)(1/2)(1/2)(1/2) = 1/16</p><p><strong>Step 3:</strong> Total probability = P(WBWB) + P(BWBW) = 1/16 + 1/16 = 2/16 = 1/8</p><p><strong>Clarification:</strong> If the question specifically asks for exactly 4 draws with alternating colors starting with a specific color, the answer is 1/16. If asking for any alternating sequence in 4 draws, it is 1/8. The given answer of 1/16 suggests the question specifies one particular alternating pattern (e.g., 'starting with white' or implicitly WBWB).</p><p>∴ Answer: 1/16</p>
Correct Answer: 1/16