Q78 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Mathematics

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

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

Solution

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 $=p\left(\frac{E_2}{A}\right)$ $=\frac{p\left(E_2\right) p\left(\frac{A}{E_2}\right)}{p\left(E_1\right) p\left(\frac{A}{E_1}\right)+p\left(E_2\right) p\left(\frac{A}{E_2}\right)}=\frac{\frac{4}{7} \times \frac{64}{27}}{\frac{3}{7} \times \frac{5}{16}+\frac{4}{7} \times \frac{64}{27}}$ $=\frac{1024}{1429}$

Hence, the correct answer is option: 'C'.

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