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 3/4 and the remaining 6 questions correctly with probability ¼. 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

Correct Answer: 479

The correct answer is 479
Student guesses only two wrong. So there are only three possibilities
(a) Student guesses both wrong from 1^st section
(b) Student guesses both wrong from 2^nd section
(c) Student guesses two wrong one from each section
Required probabilities
\(= 4C_2\left(\frac{3}{4}\right)^2\left(\frac{1}{4}\right)^2\left(\frac{1}{4}\right)^6 + 6C_2\left(\frac{3}{4}\right)^2\left(\frac{1}{4}\right)^4\left(\frac{1}{4}\right)^4 + 4C_1\cdot 6C_1\left(\frac{3}{4}\right)\left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^3\left(\frac{1}{4}\right)^5\)
\(=\frac{1}{4^{10}}\left[6 \times 9 + 15 \times 9^4 + 24 \times 9^2\right]\)

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

\(=\frac{27 \times 479}{4^{10}}\)