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 

Student guesses only two wrong. So there are three possibilities.
(i) Student guesses both wrong from 1st section
(ii) Student guesses both wrong from 2nd section
(iii) Student guesses two wrong one from each section Required probabilities
$\begin{array}{l}{=}^4{C}_2{\left(\frac{3}{2}\right)}^2{\left(\frac{1}{4}\right)}^2{\left(\frac{1}{6}\right)}^6{+}^6{C}_2{\left(\frac{3}{4}\right)}^2{\left(\frac{1}{4}\right)}^4{\left(\frac{3}{4}\right)}^4{+}^4{C}_1{.}^6{C}_1\left(\frac{3}{4}\right)\left(\frac{1}{4}\right){\left(\frac{3}{4}\right)}^3{\left(\frac{1}{4}\right)}^5\\ =\frac{1}{4^{10}}[6\times 9+15\times 9^4+24\times 9^2]\\ =\frac{27}{4^{10}}[2+27\times 15+72]\\ =\frac{27\times 479}{4^{10}}\end{array}$