In a test an examinee either guesses or copies or knows the answer to

1.	In a test an examinee either guesses or copies or knows the answer to a multiple choice question with four choices. The probability that he makes a guess is \(\frac{1}{3}\) and the probability that he copies the answer is \(\frac{1}{6}\) The probability that his answer is correct given that he copied it is \(\frac{1}{8}\) The probability that his answer is correct, given that he guessed it is \(\frac{1}{4}\) The probability that they knew the answer to the questions given that he correctly answered it is (a) \(\frac{24}{31}\) (b) \(\frac{31}{24}\) (c) \(\frac{24}{29}\) (d) \(\frac{29}{24}\)
Answer» Answer : (c) \(\frac{24}{29}\) Let, E1: Examinee guesses the answer E2: Examinee copies the answer E3: Examinee knows the answer E : Event examinee answers correctly Given, P(E₁) = \(\frac{1}{3}\) , P(E₂) = \(\frac{1}{6}\) ∴ P(E₃) = 1 – (P(E₁) + P(E₃)) = 1- \(\big(\frac{1}{3}+\frac{1}{6}\big)\) = \(\frac{1}{2}\) Given, \(P(E/E_1) =\frac{1}{4},P(E/E_2) =\frac{1}{8},P(E/E_3)=1\) ∴ Required probability = P(E₃/E) = \(\frac{P(E_3).P(E/E_3)}{P(E_1).P(E/E_1)+P(E_2).P(E/E_2)+P(E_3).P(E/E_3)} \) \(= \frac{\frac{1}{2}.1}{\frac{1}{3}\times \frac{1}{4}+\frac{1}{6}\times \frac{1}{8}+\frac{1}{2}\times 1}\) = \(\frac{\frac{1}{2}}{\frac{1}{12}+\frac{1}{48}+\frac{1}{2}}\) = \(\frac{\frac{1}{2}}{\frac{4+1+24}{48}}\) = \(\frac{\frac{1}{2}}{\frac{29}{48}}\) \(=\frac{24}{29}\)

In a test an examinee either guesses or copies or knows the answer to a multiple choice question with four choices. The probability that he makes a guess is \(\frac{1}{3}\) and the probability that he copies the answer is \(\frac{1}{6}\) The probability that his answer is correct given that he copied it is \(\frac{1}{8}\) The probability that his answer is correct, given that he guessed it is \(\frac{1}{4}\) The probability that they knew the answer to the questions given that he correctly answered it is (a) \(\frac{24}{31}\) (b) \(\frac{31}{24}\) (c) \(\frac{24}{29}\) (d) \(\frac{29}{24}\)

Answer»

Answer : (c) \(\frac{24}{29}\)

Let,

E1: Examinee guesses the answer

E2: Examinee copies the answer

E3: Examinee knows the answer

E : Event examinee answers correctly

Given, P(E₁) = \(\frac{1}{3}\) , P(E₂) = \(\frac{1}{6}\)

∴ P(E₃) = 1 – (P(E₁) + P(E₃)) = 1- \(\big(\frac{1}{3}+\frac{1}{6}\big)\) = \(\frac{1}{2}\)

Given, \(P(E/E_1) =\frac{1}{4},P(E/E_2) =\frac{1}{8},P(E/E_3)=1\)

∴ Required probability = P(E₃/E)

= \(\frac{P(E_3).P(E/E_3)}{P(E_1).P(E/E_1)+P(E_2).P(E/E_2)+P(E_3).P(E/E_3)} \)

\(= \frac{\frac{1}{2}.1}{\frac{1}{3}\times \frac{1}{4}+\frac{1}{6}\times \frac{1}{8}+\frac{1}{2}\times 1}\) = \(\frac{\frac{1}{2}}{\frac{1}{12}+\frac{1}{48}+\frac{1}{2}}\) = \(\frac{\frac{1}{2}}{\frac{4+1+24}{48}}\) = \(\frac{\frac{1}{2}}{\frac{29}{48}}\) \(=\frac{24}{29}\)

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