Answer: We can solve this problem using the binomial distribution. Let X be the number of threes obtained in 7 rolls of a fair die. Then X has a binomial distribution with n = 7 and p = 1/6, since each roll is an independent trial with a success probability of 1/6 (i.e., rolling a three).
The probability of getting at least 6 threes is equal to the sum of the probabilities of getting 6, 7 threes. That is:
P(X ≥ 6) = P(X = 6) + P(X = 7)
Using the binomial probability formula, we get:
P(X = k) = (n choose k) * p^k * (1 - p)^(n-k)
where (n choose k) is the binomial coefficient, which can be computed as:
(n choose k) = n! / (k! * (n-k)!)
where n! is the factorial of n (i.e., the product of all positive integers up to n).
Plugging in the values, we get:
P(X = 6) = (7 choose 6) * (1/6)^6 * (5/6)^1 = 0.000024
P(X = 7) = (7 choose 7) * (1/6)^7 * (5/6)^0 = 0.000001
Therefore, the probability of getting at least 6 threes is:
P(X ≥ 6) = P(X = 6) + P(X = 7) = 0.000025 (rounded to the nearest thousandth)
So the probability, rounded to the nearest thousandth, of getting at least 6 threes is 0.000025.