If the mean and the variance of a binomial variate $$X$$ are 2 and 1 respectively, then the probability that $$X$$ takes a value greater than or equal to one is:

Given that the mean and variance of a binomial variate $$X$$ are 2 and 1 respectively, we need to find the probability that $$X$$ is greater than or equal to one.

For a binomial distribution, the mean $$\mu$$ is given by $$\mu = n \cdot p$$, and the variance $$\sigma^2$$ is given by $$\sigma^2 = n \cdot p \cdot (1 - p)$$, where $$n$$ is the number of trials and $$p$$ is the probability of success.

We have:

Mean: $$n \cdot p = 2$$ ...(1)

Variance: $$n \cdot p \cdot (1 - p) = 1$$ ...(2)

Substituting equation (1) into equation (2):

$$2 \cdot (1 - p) = 1$$

Solving for $$p$$:

Divide both sides by 2: $$1 - p = \frac{1}{2}$$

Then, $$p = 1 - \frac{1}{2} = \frac{1}{2}$$

Now substitute $$p = \frac{1}{2}$$ back into equation (1):

$$n \cdot \frac{1}{2} = 2$$

Multiply both sides by 2: $$n = 4$$

So, the binomial distribution has parameters $$n = 4$$ and $$p = \frac{1}{2}$$.

We need $$P(X \geq 1)$$. This is equal to $$1 - P(X = 0)$$, since $$X$$ is non-negative.

The probability mass function for binomial distribution is:

$$P(X = k) = \binom{n}{k} \cdot p^k \cdot (1 - p)^{n - k}$$

For $$k = 0$$:

$$P(X = 0) = \binom{4}{0} \cdot \left(\frac{1}{2}\right)^0 \cdot \left(1 - \frac{1}{2}\right)^{4 - 0}$$

$$\binom{4}{0} = 1$$, $$\left(\frac{1}{2}\right)^0 = 1$$, and $$\left(\frac{1}{2}\right)^4 = \frac{1}{16}$$

So, $$P(X = 0) = 1 \cdot 1 \cdot \frac{1}{16} = \frac{1}{16}$$

Therefore,

$$P(X \geq 1) = 1 - P(X = 0) = 1 - \frac{1}{16} = \frac{16}{16} - \frac{1}{16} = \frac{15}{16}$$

Hence, the probability that $$X$$ takes a value greater than or equal to one is $$\frac{15}{16}$$.

Comparing with the options:

A. $$\frac{1}{16}$$

B. $$\frac{9}{16}$$

C. $$\frac{3}{4}$$

D. $$\frac{15}{16}$$

So, the correct answer is Option D.