If the sum and the product of mean and variance of a binomial distribution are $$24$$ and $$128$$ respectively, then the probability of one or two successes is

For a binomial distribution with parameters $$n$$ and $$p$$, the mean is $$np$$ and the variance is $$npq$$ where $$q = 1 - p$$.

Since the mean plus variance equals 24 and their product is 128, we have

$$np + npq = 24 \implies np(1 + q) = 24$$

$$np \cdot npq = 128 \implies n^2p^2q = 128$$

To simplify these equations, introduce the notation $$m = np$$, which transforms the relationships into

$$m(1 + q) = 24 \quad \ldots (1)$$

$$m^2 q = 128 \quad \ldots (2)$$

From (1), it follows that

$$m = \dfrac{24}{1 + q}$$

Substituting this expression for $$m$$ into equation (2) yields:

$$\dfrac{576}{(1+q)^2} \cdot q = 128$$

$$576q = 128(1+q)^2 = 128(1 + 2q + q^2)$$

$$576q = 128 + 256q + 128q^2$$

$$128q^2 - 320q + 128 = 0$$

$$q^2 - 2.5q + 1 = 0 \implies 2q^2 - 5q + 2 = 0$$

$$(2q - 1)(q - 2) = 0$$

$$q = \dfrac{1}{2} \text{ or } q = 2$$

Since $$0 \le q \le 1$$, the valid solution is $$q = \dfrac{1}{2}$$ and hence $$p = \dfrac{1}{2}$$.

Having determined $$p$$, the relation $$m = np$$ gives

$$m = np = \dfrac{24}{1 + \frac{1}{2}} = \dfrac{24}{\frac{3}{2}} = 16$$

$$n = \dfrac{16}{p} = \dfrac{16}{1/2} = 32$$

Finally, the probability of observing one or two successes is computed as

$$P(X = 1) + P(X = 2) = \binom{32}{1}\left(\dfrac{1}{2}\right)^{32} + \binom{32}{2}\left(\dfrac{1}{2}\right)^{32}$$

$$= \dfrac{1}{2^{32}}\left(32 + \dfrac{32 \cdot 31}{2}\right) = \dfrac{1}{2^{32}}(32 + 496) = \dfrac{528}{2^{32}}$$

$$= \dfrac{528}{2^{32}} = \dfrac{33 \times 16}{2^{32}} = \dfrac{33 \times 2^4}{2^{32}} = \dfrac{33}{2^{28}}$$

Thus, the required probability is $$\dfrac{33}{2^{28}}$$, which corresponds to Option C.