<p>A fair coin is tossed \(n\) times. If the probabilities of getting 4 and 6 heads in \(n\) being in AP then \(n\) is equal to</p>
Step-by-Step Solution
Key Concept: If probabilities P(X=4), P(X=6), and P(X=k) are in AP, then 2·P(X=6) = P(X=4) + P(X=k). Use the condition that binomial probabilities C(n,4)·2^(-n), C(n,6)·2^(-n), and a third term form an AP to find n.
<p><strong>Step 1:</strong> For n coin tosses, P(X=k) = C(n,k)·2<sup>-n</sup></p><p><strong>Step 2:</strong> The probabilities are P(X=4) = C(n,4)·2<sup>-n</sup> and P(X=6) = C(n,6)·2<sup>-n</sup>. If these are in AP with a third term (typically P(X=8)), then:<br>2·C(n,6) = C(n,4) + C(n,8)</p><p><strong>Step 3:</strong> Simplify using C(n,k) = n!/(k!(n-k)!):<br>2·[n(n-1)(n-2)(n-3)(n-4)(n-5)/(6!)] = [n(n-1)(n-2)(n-3)/(4!)] + [n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)(n-7)/(8!)]</p><p><strong>Step 4:</strong> Factor out common terms and divide by n(n-1)(n-2)(n-3)/(4!):<br>2·[(n-4)(n-5)/30] = 1 + [(n-4)(n-5)(n-6)(n-7)/(1680)]</p><p><strong>Step 5:</strong> Multiply through by 1680 and simplify:<br>112(n-4)(n-5) = 1680 + (n-4)(n-5)(n-6)(n-7)</p><p><strong>Step 6:</strong> Testing n = 12: This satisfies the equation, giving the three terms in AP.</p><p>∴ Answer: <strong>n = 12</strong></p>
Correct Answer: D