1. Let the number of tosses be $n$.

2. We want: $P(\text{at least one head}) \ge 0.9$

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

4. For a fair coin, probability of tail on one toss is $\frac{1}{2}$. So probability of all tails in $n$ tosses is: $\left(\frac{1}{2}\right)^n$

Hence, $1 - \left(\frac{1}{2}\right)^n \ge 0.9$

5. Solve the inequality: $\left(\frac{1}{2}\right)^n \le 0.1$

Now check integer values of $n$:

• For $n=3$: $\left(\frac{1}{2}\right)^3 = \frac{1}{8} = 0.125 > 0.1$ So 3 tosses are not enough.

• For $n=4$: $\left(\frac{1}{2}\right)^4 = \frac{1}{16} = 0.0625 < 0.1$ So 4 tosses are enough.

6. Therefore, the minimum number of tosses required is: $\boxed{4}$

7. Option check:

• A: 2 $\Rightarrow 1-\frac14=0.75$ ❌
• B: 3 $\Rightarrow 1-\frac18=0.875$ ❌
• C: 4 $\Rightarrow 1-\frac1{16}=0.9375$ ✅
• D: 5 is sufficient, but not minimum ❌

So the correct option is C.