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

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answer is 479.

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Detailed Solution

$A = {1, 2, 3, 4} : P (A) = \frac{3}{4}$

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

E=Event that the student guesses the answers of exactly 8 questions correctly out of 10 .

$(4, 4) :^{4} C_{4} {(\frac{3}{4})}^{4} \cdot^{6} C_{4} \cdot {(\frac{1}{4})}^{4} \cdot {(\frac{3}{4})}^{2}$

$(3, 5) :^{4} C_{3} {(\frac{3}{4})}^{3} \cdot {(\frac{1}{4})}^{1} \cdot^{6} C_{5} {(\frac{1}{4})}^{5} \cdot (\frac{3}{4})$

$(2, 6) :^{4} C_{2} {(\frac{3}{4})}^{2} {(\frac{1}{4})}^{2} \cdot^{6} C_{6} {(\frac{1}{4})}^{6}$

$P(E) = \frac{1}{4^{10}} [3^{4} \times 15 \times 3^{2} + 4 \times 3^{3} \times 6 \times 3 + 6 \times 3^{2}]$

$\begin{aligned} = \frac{27}{4^{10}} [27 \times 15 + 72 + 2] = \frac{27}{4^{10}} (479) \\ \Rightarrow K = 479 \end{aligned}$

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