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Home » In answering a question on a multiple choice test, a student either knows the answer or guesses. Let be the probability that he knows the answer and be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability . What is the probability that the student knows the answer given that he answered it correctly?
In answering a question on a multiple choice test, a student either knows the answer or guesses. Let 3 / 4 be the probability that he knows the answer and 1 / 4 be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability 1 / 4. What is the probability that the student knows the answer given that he answered it correctly?
CBSE | Class 12 | Guides | Maths | NCERT | Probability
In answering a question on a multiple choice test, a student either knows the answer or guesses. Let 3 / 4 be the probability that he knows the answer and 1 / 4 be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability 1 / 4. What is the probability that the student knows the answer given that he answered it correctly?

Solution:

Let E_{1} represent the situation in which the student knows the solution, E_{2} represent the situation in which the student guesses the answer, and \mathrm{A} represent the situation in which the answer is accurate.

Then P\left(E_{1}\right)=3 / 4

And \mathrm{P}\left(\mathrm{E}_{2}\right)=1 / 4

Also \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{1}\right)=\mathrm{P} (correct answer given that he knows) =1

And \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{2}\right)=\mathrm{P} (correct answer given that he guesses) =1 / 4

Now, the probability that he knows the answer is \mathrm{P} if the answer is accurate.

\left(\mathrm{E}_{1} \mid \mathrm{A}\right)

By using Bayes’ theorem, we have:

\mathrm{P}\left(\mathrm{E}_{1} \mid \mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{1}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \cdot \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{2}\right)}

We get the following results by substituting the values:

=\frac{\frac{3}{4} \cdot 1}{\frac{3}{4} \cdot 1+\frac{1}{4} \cdot \frac{1}{4}}

=\frac{\frac{3}{4}}{\frac{3}{4}+\frac{1}{16}}

=\frac{\frac{3}{4}}{\frac{13}{16}}=\frac{12}{13}

\Rightarrow \mathrm{P}\left(\mathrm{E}_{1} \mid \mathrm{A}\right)=\frac{12}{13}

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