This problem can be solved using the binomial probability distribution, as each question is an independent trial with two possible outcomes (correct or incorrect).

1. Identify the parameters for the binomial distribution:
Number of trials (questions), \(n = 10\).
Probability of success (guessing a True/False question correctly), \(p = \frac{1}{2}\).
Probability of failure (guessing incorrectly), \(q = 1 - p = \frac{1}{2}\).

2. Define the problem:
We need to find the probability of getting "at least six" correct answers. Let \(X\) be the number of correct answers. We must calculate \(P(X \ge 6)\), which is the sum of the probabilities of getting exactly 6, 7, 8, 9, or 10 correct answers.
\[P(X \ge 6) = P(X=6) + P(X=7) + P(X=8) + P(X=9) + P(X=10)\]
3. Use the Binomial Probability Formula:
The formula is \(P(X=k) = C(n, k) p^k q^{n-k}\).
Since \(p=q=\frac{1}{2}\), the formula simplifies to \(P(X=k) = C(n, k) (\frac{1}{2})^n\).
For \(n=10\), we have \(P(X=k) = C(10, k) (\frac{1}{2})^{10} = \frac{C(10, k)}{1024}\).

4. Calculate the required combinations \(C(10,k)\):
We need to calculate the number of ways to get k successes in 10 trials.
\[C(10, 6) = \frac{10!}{6!4!} = 210\]\[C(10, 7) = \frac{10!}{7!3!} = 120\]\[C(10, 8) = \frac{10!}{8!2!} = 45\]\[C(10, 9) = \frac{10!}{9!1!} = 10\]\[C(10, 10) = \frac{10!}{10!0!} = 1\]
5. Sum the probabilities:
Now we sum these outcomes and divide by the total number of possible outcomes (\(2^{10} = 1024\)).
\[\begin{aligned} P(X \ge 6) &= \frac{C(10, 6) + C(10, 7) + C(10, 8) + C(10, 9) + C(10, 10)}{1024} \\ &= \frac{210 + 120 + 45 + 10 + 1}{1024} \\ &= \frac{386}{1024} \end{aligned}\]
6. Convert to a decimal:
\[P(X \ge 6) \approx 0.37695\]
Rounding to three decimal places gives 0.377.

Result:
The probability of guessing correctly at least six answers is approximately 0.377.