Q82 / 125 · BITSAT 2018 · Mathematics

In a test an examiner either guesses or copies or knows the answer to a multiple choice question with 4 choices. The probability that he/she makes a guess is 13. The probability that he/she copies the answer is 16. If the probability that the answer is correct, given that he/she copied it is 18, then the probability that he/she knows the answer to a question given that he/she correctly answered it, is

  • A
    2729
  • B
    2629
  • C
    2529
  • Correct answer
    D
    2429

Solution

Let A be the event that the examinee gives the correct answer.

Let G, C, K stand for guessing, copying and knowing. respectively.

Given, PC=16, PG=13

and PAC=18

⇒PC+PG+PK=1

⇒16+13+PK=1

∴ PK=12

Also, PAK=1, for if the examinee knows, he/she will correctly answer it and PAG=14,

 since there are four choices.

Now, total probability PA=PGPAG+PCPAC+PKPAK

=13·14+16·18+12·1

=112+148+12=4+1+2448=2948

Now, PKA=PKPAKPA

=12·12948=12·4829=2429

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

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