The mean and the variance of a binomial distribution are $$4$$ and $$2$$ respectively. Then the probability of $$2$$ successes is

For a binomial distribution with $$ n $$ trials, probability of success $$ p $$, and probability of failure $$ q $$, the formulas for mean and variance are:

$$ \text{Mean} = np $$

$$ \text{Variance} = npq $$

We are given:

$$ np = 4 $$

$$ npq = 2 $$

Step 1: Find the Probabilities $$ p $$ and $$ q $$

To find $$ q $$, we divide the variance by the mean:

$$ \frac{npq}{np} = \frac{2}{4} $$

$$ q = \frac{1}{2} $$

Since the sum of the probabilities of success and failure is always 1 ($$ p + q = 1 $$), we can find $$ p $$:

$$ p = 1 - q = 1 - \frac{1}{2} = \frac{1}{2} $$

Step 2: Find the Number of Trials $$ n $$

Substitute the value of $$ p $$ back into the mean equation:

$$ n \cdot \left(\frac{1}{2}\right) = 4 $$

$$ n = 8 $$

Step 3: Calculate the Probability of 2 Successes

The probability of getting exactly $$ k $$ successes in a binomial distribution is given by the formula:

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

We need to find the probability for $$ k = 2 $$ successes:

$$ P(X = 2) = \binom{8}{2} \cdot \left(\frac{1}{2}\right)^2 \cdot \left(\frac{1}{2}\right)^{8-2} $$

$$ P(X = 2) = \binom{8}{2} \cdot \left(\frac{1}{2}\right)^2 \cdot \left(\frac{1}{2}\right)^6 $$

$$ P(X = 2) = \binom{8}{2} \cdot \left(\frac{1}{2}\right)^8 $$

Now evaluate the combination and power terms:

$$ \binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28 $$

$$ \left(\frac{1}{2}\right)^8 = \frac{1}{256} $$

Substitute these values back into the equation:

$$ P(X = 2) = 28 \cdot \frac{1}{256} $$

$$ P(X = 2) = \frac{28}{256} $$

Therefore, the probability of 2 successes is $$ \frac{28}{256} $$.