Question:
<p>A fair coin is tossed repeatedly. The minimum number of tosses \(N\) required so that \(P(\text{at least one head}) > \dfrac{63}{64}\) is <em>[JEE Advanced 2013]</em></p>
Step-by-Step Solution
Key Concept: P(at least one head in n tosses) = 1-(1/2)ⁿ > 63/64 ⟺ (1/2)ⁿ < 1/64 = (1/2)^6 ⟺ n \geq 6.
<p>$P(\text{at least one head}) = 1 - \left(\frac{1}{2}\right)^N > \frac{63}{64} = 1 - \frac{1}{64}$.</p><p>$\left(\frac{1}{2}\right)^N < \frac{1}{64} = \left(\frac{1}{2}\right)^6$.</p><p>So $N \geq 6$. Minimum $N = 6$.</p>
Correct Answer: 6