Let X have a binomial distribution $$B(n, p)$$ such that the sum and the product of the mean and variance of X are 24 and 128 respectively. If $$P(X > n-3) = \frac{k}{2^n}$$, then $$k$$ is equal to

We have $$X \sim B(n, p)$$ with mean $$\mu = np$$ and variance $$\sigma^2 = npq$$ (where $$q = 1 - p$$). We are given:

$$\mu + \sigma^2 = 24$$ and $$\mu \cdot \sigma^2 = 128$$.

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

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

From the first equation: $$np = \frac{24}{1+q} = \frac{24}{2-p}$$.

From the second: $$(np)^2 \cdot q = 128$$.

Substituting: $$\left(\frac{24}{2-p}\right)^2 (1-p) = 128$$.

$$\frac{576(1-p)}{(2-p)^2} = 128$$

$$576(1-p) = 128(2-p)^2$$

$$576 - 576p = 128(4 - 4p + p^2)$$

$$576 - 576p = 512 - 512p + 128p^2$$

$$64 - 64p = 128p^2$$

$$1 - p = 2p^2$$

$$2p^2 + p - 1 = 0$$

$$(2p - 1)(p + 1) = 0$$

So $$p = \frac{1}{2}$$ (rejecting $$p = -1$$).

Then $$q = \frac{1}{2}$$, and $$np(1 + q) = np \cdot \frac{3}{2} = 24$$, so $$np = 16$$, giving $$n = 32$$.

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

$$P(X > 29) = \frac{1}{2^{32}}\left[\binom{32}{30} + \binom{32}{31} + \binom{32}{32}\right]$$

$$\binom{32}{30} = \binom{32}{2} = \frac{32 \times 31}{2} = 496$$

$$\binom{32}{31} = 32$$

$$\binom{32}{32} = 1$$

$$P(X > 29) = \frac{496 + 32 + 1}{2^{32}} = \frac{529}{2^{32}}$$

Since $$P(X > n - 3) = \frac{k}{2^n} = \frac{k}{2^{32}}$$, we get $$k = 529$$.

The correct answer is Option B: 529.