Question

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 $\frac{2}{3} .$ The probability that a student guesses a correct answer is $\frac{1}{2} .$ 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 $\frac{3}{7}$. Given that a student answered at least $3$ questions correctly, the probability that he answered all the questions without guessing is

• A. $\frac{13}{15}$
• B. $\frac{405}{1429}$
• C. $\frac{1024}{1429}$
• D. $\frac{2}{15}$

Answer 1

Answer: (C)

Consider the events
$E_{1}=$ Answer by student without guessing
$E_{2}=$ Answer by student guessing
$A=$ At least three question correctly
$P\left(E_{1}\right)=\frac{3}{7}, p\left(B_{2}\right)=\frac{4}{7}$
$p\left(\frac{A}{E_{1}}\right)={ }^{4} C_{3}\left(\frac{1}{2}\right)^{4}+{ }^{4} C_{4}\left(\frac{1}{2}\right)^{4}=\left(\frac{1}{2}\right)^{4}(4+1)=\frac{5}{16}$
$p\left(\frac{A}{E_{2}}\right)={ }^{4} C_{3}\left(\frac{2}{3}\right)^{3} \times \frac{1}{3}+{ }^{4} C_{4}\left(\frac{2}{3}\right)^{4}=\left(\frac{2}{3}\right)^{3}\left(\frac{4}{3}+\frac{2}{3}\right)$
$p\left(\frac{A}{E_{2}}\right)=\frac{16}{27}$
Required probability
$=p\left(\frac{E_{2}}{A}\right)$
$=\frac{p\left(E_{2}\right) p\left(\frac{A}{E_{2}}\right)}{p\left(E_{1}\right) p\left(\frac{A}{E_{1}}\right)+p\left(E_{2}\right) p\left(\frac{A}{E_{2}}\right)}$
$=\frac{\frac{4}{7} \times \frac{64}{27}}{\frac{3}{7} \times \frac{5}{16}+\frac{4}{7} \times \frac{64}{27}}=\frac{1024}{1429}$