In an examination, there are 10 true-false type questions. Out of 10 , a student can guess the answer of 4…

Given,

$A=\left\{1,2,3,4\right\}:P\left(A\right)=\frac{3}{4}\to$Correct answer

$B=\left\{5,6,7,8,9,10\right\};P\left(B\right)=\frac{1}{4}$ Correct answer

Now total ways in which $8$ Correct answer is given will be,

$\Rightarrow \text{Case (i) }\left(4,4\right):{}^{4}C_4{\left(\frac{3}{4}\right)}^4·{}^{6}C_4·{\left(\frac{1}{4}\right)}^4·{\left(\frac{3}{4}\right)}^2$

$\Rightarrow \mathrm{Case} \text{(ii) }\left(3,5\right):{}^{4}C_3{\left(\frac{3}{4}\right)}^3·{\left(\frac{1}{4}\right)}^1·{}^{6}C_5{\left(\frac{1}{4}\right)}^5·\left(\frac{3}{4}\right)$

$\Rightarrow \text{Case (iii)} \left(2,6\right):{}^{4}C_2{\left(\frac{3}{4}\right)}^2{\left(\frac{1}{4}\right)}^2·{}^{6}C_6{\left(\frac{1}{4}\right)}^6$

Total probability  $=\frac{1}{4^{10}}\left[3^4\times 15\times 3^2+4\times 3^3\times 6\times 3+6\times 3^2\right]$

$=\frac{27}{4^{10}}\left[2.7\times 15+72+2\right]$

Now comparing with $\frac{27k}{4^{10}}$

We get,$\mathrm{K}=479$