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 ${3 \over 4}$ and the remaining 6 questions correctly with probability ${1 \over 4}$. If the probability that the student guesses the answers of exactly 8 questions correctly out of 10 is ${{{{27}k}} \over {{4^{10}}}}$, then k is equal to ___________.

<p>Student guesses only two wrong. So there are three possibilities.</p> <p>(i) Student guesses both wrong from 1^st section</p> <p>(ii) Student guesses both wrong from 2^nd section</p> <p>(iii) Student guesses two wrong one from each section</p> <p>Required probabilities</p> <p>$$ = {}^4{C_2}{\left( {{3 \over 4}} \right)^2}{\left( {{1 \over 4}} \right)^2}{\left( {{1 \over 6}} \right)^6} + {}^6{C_2}{\left( {{3 \over 4}} \right)^2}{\left( {{1 \over 4}} \right)^4}{\left( {{3 \over 4}} \right)^4} + {}^4{C_1}\,.\,{}^6{C_1}\left( {{3 \over 4}} \right)\left( {{1 \over 4}} \right){\left( {{3 \over 4}} \right)^3}{\left( {{1 \over 4}} \right)^5}$$</p> <p>$$ = {1 \over {{4^{10}}}}\left[ {6 \times 9 + 15 \times {9^4} + 24 \times {9^2}} \right]$$</p> <p>$= {{27} \over {{4^{10}}}}\left[ {2 + 27 \times 15 + 72} \right]$</p> <p>$= {{27 \times 479} \over {{4^{10}}}}$</p>