Question

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 $\frac{27 k }{4^{10}}$, then $k$ is equal to ______.

Answer 1

$A =\{1,2,3,4\}: P ( A )=\frac{3}{4} \rightarrow $ Correct
$B =\{5,6,7,8,9,10\} ; P ( B )=\frac{1}{4}$ Correct
$8$ Correct Ans.:
$(4,4):{ }^{4} C _{4}\left(\frac{3}{4}\right)^{4} \cdot{ }^{6} C _{4} \cdot\left(\frac{1}{4}\right)^{4} \cdot\left(\frac{3}{4}\right)^{2}$
$(3,5):{ }^{4} C _{3}\left(\frac{3}{4}\right)^{3} \cdot\left(\frac{1}{4}\right)^{1} \cdot{ }^{6} C _{5}\left(\frac{1}{4}\right)^{5} \cdot\left(\frac{3}{4}\right)$
$(2,6):{ }^{4} C _{2}\left(\frac{3}{4}\right)^{2}\left(\frac{1}{4}\right)^{2} \cdot{ }^{6} C _{6}\left(\frac{1}{4}\right)^{6}$
Total $=\frac{1}{4^{10}}\left[3^{4} \times 15 \times 3^{2}+4 \times 3^{3} \times 6 \times 3+6 \times 3^{2}\right]$
$=\frac{27}{4{ }^{10}}[2.7 \times 15+72+2]$
$\Rightarrow K =479$