Search any question & find its solution
Question:
Answered & Verified by Expert
In an examination there are four Yes/No type of questions. The probability that the answer by the student to a question without guess to be correct is $2 / 3$. The probability that a student guesses a correct answer is $1 / 2$. A student writes the examination either by without guessing answers to all the 4 questions or by guessing answers to all 4 questions. The probability that he attempt the exam by guessing answers to all questions is $3 / 7$. Given that a student answered at least 3 questions correctly, the probability that he answered all the questions without guessing is
Options:
Solution:
2369 Upvotes
Verified Answer
The correct answer is:
$\frac{1024}{1429}$
Consider the events
$E_1=$ Answer by student without guessing
$E_2=$ Answer by student guessing
$A=$ At least three question correctly
$P\left(E_1\right)=\frac{3}{7}, p\left(E_2\right)=\frac{4}{7}$
$p\left(\frac{A}{E_1}\right)={ }^4 C_3\left(\frac{1}{2}\right)^4+{ }^4 C_4\left(\frac{1}{2}\right)^4=\left(\frac{1}{2}\right)^4(4+1)=\frac{5}{16}$
$p\left(\frac{A}{E_2}\right)={ }^4 C_3\left(\frac{2}{3}\right)^3 \times \frac{1}{3}+{ }^4 C_4\left(\frac{2}{3}\right)^4=\left(\frac{2}{3}\right)^3\left(\frac{4}{3}+\frac{2}{3}\right)$
$p\left(\frac{A}{E_2}\right)=\frac{16}{27}$
Required probability $=p\left(\frac{E_2}{A}\right)$
$=\frac{p\left(E_2\right) p\left(\frac{A}{E_2}\right)}{p\left(E_1\right) p\left(\frac{A}{E_1}\right)+p\left(E_2\right) p\left(\frac{A}{E_2}\right)}=\frac{\frac{4}{7} \times \frac{64}{27}}{\frac{3}{7} \times \frac{5}{16}+\frac{4}{7} \times \frac{64}{27}}$
$=\frac{1024}{1429}$
$E_1=$ Answer by student without guessing
$E_2=$ Answer by student guessing
$A=$ At least three question correctly
$P\left(E_1\right)=\frac{3}{7}, p\left(E_2\right)=\frac{4}{7}$
$p\left(\frac{A}{E_1}\right)={ }^4 C_3\left(\frac{1}{2}\right)^4+{ }^4 C_4\left(\frac{1}{2}\right)^4=\left(\frac{1}{2}\right)^4(4+1)=\frac{5}{16}$
$p\left(\frac{A}{E_2}\right)={ }^4 C_3\left(\frac{2}{3}\right)^3 \times \frac{1}{3}+{ }^4 C_4\left(\frac{2}{3}\right)^4=\left(\frac{2}{3}\right)^3\left(\frac{4}{3}+\frac{2}{3}\right)$
$p\left(\frac{A}{E_2}\right)=\frac{16}{27}$
Required probability $=p\left(\frac{E_2}{A}\right)$
$=\frac{p\left(E_2\right) p\left(\frac{A}{E_2}\right)}{p\left(E_1\right) p\left(\frac{A}{E_1}\right)+p\left(E_2\right) p\left(\frac{A}{E_2}\right)}=\frac{\frac{4}{7} \times \frac{64}{27}}{\frac{3}{7} \times \frac{5}{16}+\frac{4}{7} \times \frac{64}{27}}$
$=\frac{1024}{1429}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.