In an examination, there are $$10$$ true-false type questions. Out of $$10$$, a student can guess the answer of $$4$$ questions correctly with probability $$\frac{3}{4}$$ and the remaining $$6$$ questions correctly with probability $$\frac{1}{4}$$. If the probability that the student guesses the answers of exactly $$8$$ questions correctly out of $$10$$ is $$\frac{27k}{4^{10}}$$, then $$k$$ is equal to ______.

There are 10 true-false questions. For 4 of them, the student guesses correctly with probability $$\frac{3}{4}$$, and for the remaining 6, the probability of a correct guess is $$\frac{1}{4}$$.

We need $$P(\text{exactly 8 correct out of 10})$$.

Let $$X$$ = number of correct answers among the 4 questions (probability $$\frac{3}{4}$$ each).

Let $$Y$$ = number of correct answers among the 6 questions (probability $$\frac{1}{4}$$ each).

We need $$P(X + Y = 8) = \sum_{i} P(X = i) \cdot P(Y = 8-i)$$.

Since $$X \leq 4$$ and $$Y \leq 6$$, we need $$i \geq 2$$ and $$8 - i \leq 6$$, so $$i \geq 2$$. Also $$i \leq 4$$.

Case $$i = 2$$, $$Y = 6$$:

$$P(X=2) = \binom{4}{2}\left(\frac{3}{4}\right)^2\left(\frac{1}{4}\right)^2 = 6 \cdot \frac{9}{16} \cdot \frac{1}{16} = \frac{54}{256}$$

$$P(Y=6) = \binom{6}{6}\left(\frac{1}{4}\right)^6 = \frac{1}{4096}$$

Product: $$\frac{54}{256} \cdot \frac{1}{4096} = \frac{54}{4^{10}}$$

Case $$i = 3$$, $$Y = 5$$:

$$P(X=3) = \binom{4}{3}\left(\frac{3}{4}\right)^3\left(\frac{1}{4}\right)^1 = 4 \cdot \frac{27}{64} \cdot \frac{1}{4} = \frac{108}{256}$$

$$P(Y=5) = \binom{6}{5}\left(\frac{1}{4}\right)^5\left(\frac{3}{4}\right)^1 = 6 \cdot \frac{1}{1024} \cdot \frac{3}{4} = \frac{18}{4096}$$

Product: $$\frac{108}{256} \cdot \frac{18}{4096} = \frac{1944}{4^{10}}$$

Case $$i = 4$$, $$Y = 4$$:

$$P(X=4) = \left(\frac{3}{4}\right)^4 = \frac{81}{256}$$

$$P(Y=4) = \binom{6}{4}\left(\frac{1}{4}\right)^4\left(\frac{3}{4}\right)^2 = 15 \cdot \frac{1}{256} \cdot \frac{9}{16} = \frac{135}{4096}$$

Product: $$\frac{81}{256} \cdot \frac{135}{4096} = \frac{10935}{4^{10}}$$

Total probability: $$\frac{54 + 1944 + 10935}{4^{10}} = \frac{12933}{4^{10}}$$

We are told this equals $$\frac{27k}{4^{10}}$$.

$$27k = 12933$$

$$k = \frac{12933}{27} = 479$$

The correct answer is $$479$$.