In an examination, there are 10 true-false type questions. Out of...

Given,

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

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

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

$⇒ \text{Case (i)}\left(4, 4\right): C4 4 {\left(\dfrac{3}{4}\right)}^{4} \cdot C4 6 \cdot {\left(\dfrac{1}{4}\right)}^{4} \cdot {\left(\dfrac{3}{4}\right)}^{2}$

$⇒ Case\text{(ii)}\left(3, 5\right): C3 4 {\left(\dfrac{3}{4}\right)}^{3} \cdot {\left(\dfrac{1}{4}\right)}^{1} \cdot C5 6 {\left(\dfrac{1}{4}\right)}^{5} \cdot \left(\dfrac{3}{4}\right)$

$⇒ \text{Case (iii)}\left(2, 6\right): C2 4 {\left(\dfrac{3}{4}\right)}^{2}{\left(\dfrac{1}{4}\right)}^{2} \cdot C6 6 {\left(\dfrac{1}{4}\right)}^{6}$

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

$= \dfrac{27}{{4}^{10}}\left[2 . 7 \times 15 + 72 + 2\right]$

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

We get,$K = 479$