Let a random variable X has a binomial distribution with mean 8 and variance 4. If $$P(X \leq 2) = \frac{k}{2^{16}}$$, then the value of k is equal to

We know that if a random variable $$X$$ follows a binomial distribution, we write $$X \sim \text{Binomial}(n,p)$$, where

$$\text{Mean} = np \quad\text{and}\quad \text{Variance} = npq,$$

with $$q = 1-p$$.

According to the statement, the mean equals $$8$$ and the variance equals $$4$$. So we set up the two standard relations:

$$np = 8 \qquad\text{(1)}$$

$$npq = 4 \qquad\text{(2)}$$

From equation (1) we already have $$np = 8$$. Now divide equation (2) by equation (1):

$$\frac{npq}{np} \;=\; \frac{4}{8} \;\Longrightarrow\; q = \frac{1}{2}.$$

Because $$p + q = 1$$, we immediately get

$$p = 1 - q = 1 - \frac12 = \frac12.$$

Substituting $$p = \dfrac12$$ into equation (1), we find $$n$$:

$$n \Bigl(\frac12\Bigr) = 8 \;\Longrightarrow\; n = 16.$$

Thus $$X \sim \text{Binomial}(16,\tfrac12).$$ For such a distribution each term has probability

$$P(X = r) = \binom{16}{r}\Bigl(\frac12\Bigr)^r\Bigl(\frac12\Bigr)^{16-r} = \frac{\binom{16}{r}}{2^{16}}.$$

We are asked for $$P(X \leq 2)$$, i.e. the sum of the probabilities for $$r = 0,1,2$$:

$$P(X \le 2) = P(X=0) + P(X=1) + P(X=2)$$

$$\;\;\;= \frac{\binom{16}{0}}{2^{16}} + \frac{\binom{16}{1}}{2^{16}} + \frac{\binom{16}{2}}{2^{16}}.$$

Now we compute each binomial coefficient explicitly:

$$\binom{16}{0} = 1,$$

$$\binom{16}{1} = 16,$$

$$\binom{16}{2} = \frac{16 \times 15}{2} = 120.$$

Adding them gives

$$1 + 16 + 120 = 137.$$

Hence

$$P(X \le 2) = \frac{137}{2^{16}}.$$

Comparing with the required form $$\dfrac{k}{2^{16}},$$ we identify

$$k = 137.$$

Hence, the correct answer is Option D.