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

Answer 1

Answer: 479

$A=\{1,2,3,4\}:P(A)=\frac{3}{4}\to$ 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 $=\frac{1}{4^{10}}\left[3^4\times 15\times 3^2+4\times 3^3\times 6\times 3+6\times 3^2\right]$
$=\frac{27}{4{}^{10}}[2.7\times 15+72+2]$
$\Rightarrow K=479$