To solve this, we need to find the probability of getting at least two heads when a fair coin is tossed 10 times.

The easiest way to do this is to first calculate the probability of getting 0 or 1 head, and then subtract that from 1 to get the probability of getting at least two heads.

The probability of getting exactly \( k \) heads in \( n \) tosses of a fair coin follows the binomial distribution:

\[
P(X = k) = \binom{n}{k} \times p^k \times (1 - p)^{n - k}
\]

where:
- \( n = 10 \) (number of tosses),
- \( p = 0.5 \) (probability of getting a head on any toss),
- \( k \) is the number of heads.

1. Probability of getting 0 heads:
\[
P(X = 0) = \binom{10}{0} \times (0.5)^0 \times (0.5)^{10} = 1 \times 1 \times \frac{1}{1024} = \frac{1}{1024}
\]

2. Probability of getting 1 head:
\[
P(X = 1) = \binom{10}{1} \times (0.5)^1 \times (0.5)^9 = 10 \times \frac{1}{2} \times \frac{1}{512} = \frac{10}{1024}
\]

So, the total probability of getting 0 or 1 head is:

\[
P(X = 0 \text{ or } 1) = \frac{1}{1024} + \frac{10}{1024} = \frac{11}{1024}
\]

Now, subtract this from 1 to find the probability of getting at least two heads:

\[
P(\text{at least 2 heads}) = 1 - \frac{11}{1024} = \frac{1024 - 11}{1024} = \frac{1013}{1024}
\]