The correct answer is $\boxed{{}^{10}{{\text{C}}_2}{\left( {\frac{1}{2}} \right)^2}}$.
The probability of getting heads on any given toss of a fair coin is $\frac{1}{2}$. The probability of getting tails is also $\frac{1}{2}$. The probability of getting heads on the first two tosses is $\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$.
To get the probability of getting only heads on the first two tosses and tails on the remaining eight tosses, we need to multiply the probability of getting heads on the first two tosses by the number of ways to choose two tosses from ten tosses, which is $10 \choose 2$. We also need to multiply by the probability of getting tails on each of the remaining eight tosses, which is $\frac{1}{2}$.
The probability of getting only heads on the first two tosses is therefore:
$$\frac{1}{4} \cdot {}^{10}{{\text{C}}_2} \cdot \left( \frac{1}{2} \right)^8 = {}^{10}{{\text{C}}_2}{\left( {\frac{1}{2}} \right)^2}$$
Here is a brief explanation of each option:
- Option A: $\left( {\frac{1}{2}} \right)^2$ is the probability of getting heads on two consecutive tosses. This is not the correct answer because it does not take into account the possibility of getting tails on the remaining eight tosses.
- Option B: $^{10}{{\text{C}}_2}{\left( {\frac{1}{2}} \right)^8}$ is the probability of getting tails on the remaining eight tosses, given that the first two tosses are heads. This is not the correct answer because it does not take into account the possibility of getting heads on the first two tosses and tails on the remaining eight tosses.
- Option C: $\left( {\frac{1}{2}} \right)^{10}$ is the probability of getting tails on all ten tosses. This is not the correct answer because it does not take into account the possibility of getting heads on the first two tosses.