<p>A coin is tossed 8 times. The probability of having 4 heads in the even places of tosses is</p>
<p>A. \(\left(\dfrac{1}{2}\right)^3\)</p>
<p>B. \(\left(\dfrac{1}{2}\right)^8\)</p>
<p>C. \({}^8C_4\left(\dfrac{1}{2}\right)^8\)</p>
<p>D. None of these</p>
Step-by-Step Solution
Key Concept: Even places in 8 tosses are fixed positions (2, 4, 6, 8), giving exactly 4 even places. We need all 4 even places to show heads, which is independent of odd places since each toss is independent.
<p><strong>Step 1:</strong> Identify the even positions in 8 tosses: positions 2, 4, 6, 8 (exactly 4 even places).</p><p><strong>Step 2:</strong> For 4 heads in even places means heads must appear at ALL 4 even positions (2, 4, 6, 8).</p><p><strong>Step 3:</strong> Each toss is independent with P(head) = 1/2. The odd positions (1, 3, 5, 7) can be anything.</p><p><strong>Step 4:</strong> Probability = P(H at 2) × P(H at 4) × P(H at 6) × P(H at 8) = (1/2)⁴ = 1/16</p><p><strong>Step 5:</strong> The odd positions contribute a factor of 1 (any outcome is allowed), so final probability = 1/16.</p><p>∴ Answer: <strong>1/16</strong> (Option C)</p>
Correct Answer: C