In an examination, there are 10… - JEE Main 2022 Mathematics PYQ

JEE Main 2022 — Mathematics Probability & Statistics

2022integermedium

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

Official previous-year question

Held on 24 Jun 2022 · Verified 6 Jul 2026.

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Solution

Given,

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

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

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

$\Rightarrow \text{Case (i) }(4,4):C44{(\frac{3}{4})}^{4}\cdot C46\cdot {(\frac{1}{4})}^{4}\cdot {(\frac{3}{4})}^{2}$

$\Rightarrow \mathrm{Case}\text{(ii) }(3,5):C34{(\frac{3}{4})}^{3}\cdot {(\frac{1}{4})}^{1}\cdot C56{(\frac{1}{4})}^{5}\cdot (\frac{3}{4})$

$\Rightarrow \text{Case (iii)}(2,6):C24{(\frac{3}{4})}^{2}{(\frac{1}{4})}^{2}\cdot C66{(\frac{1}{4})}^{6}$

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

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

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

We get, $K=479$

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