1. Let the coin be tossed $n$ times.

2. We need: $P(\text{at least one head}) > 0.99$

3. Use the complement event: $P(\text{at least one head}) = 1 - P(\text{no head})$

For a fair coin, $P(\text{tail}) = \frac{1}{2}$, so $P(\text{no head in } n \text{ tosses}) = \left(\frac{1}{2}\right)^n$

Thus, $1 - \left(\frac{1}{2}\right)^n > 0.99$

4. Solve the inequality: $\left(\frac{1}{2}\right)^n < 0.01$

Now check powers of $2$: $\left(\frac{1}{2}\right)^6 = \frac{1}{64} = 0.015625 > 0.01$ $\left(\frac{1}{2}\right)^7 = \frac{1}{128} = 0.0078125 < 0.01$

So $n=6$ is not enough, but $n=7$ works.

5. Therefore, the minimum number of tosses required is $\boxed{7}$

6. Option check:

• A: $6$ ❌
• B: $5$ ❌
• C: $8$ ❌ (works, but not minimum)
• D: $7$ ✅

Hence, the correct option is $\boxed{D}$.

Comparison with stored answer: Stored correct answer is $D$, which matches our result.