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Question
In answering a question on a multiple choice test a student either knows the answer or guesses. Let \[\frac{3}{4}\] be the probability that he knows the answer and \[\frac{1}{4}\] be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability \[\frac{1}{4}\]. What is the probability that a student knows the answer given that he answered it correctly?
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Solution
Let A, E1 and E2 denote the events that the answer is correct, the student knows the answer and the student guesses the answer, respectively.
\[\therefore P\left( E_1 \right) = \frac{3}{4} \]
\[ P\left( E_2 \right) = \frac{1}{4}\]
\[\text { Now} , \]
\[P\left( A/ E_1 \right) = 1\]
\[P\left( A/ E_2 \right) = \frac{1}{4}\]
\[\text{ Using Bayes' theorem, we get } \]
\[\text{ Required probability } = P\left( E_1 /A \right) = \frac{P\left( E_1 \right)P\left( A/ E_1 \right)}{P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right)}\]
\[ = \frac{\frac{3}{4} \times 1}{\frac{3}{4} \times 1 + \frac{1}{4} \times \frac{1}{4}}\]
\[ = \frac{3}{3 + \frac{1}{4}} = \frac{12}{13}\]
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Solution: Let A, C and T be the events that Mr. X goes to office by Auto, Car and Train respectively. Let L be event that he is late.
Given that P(A) = `square`, P(C) = `square`
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`= square`
`"P"("C"//"L") = ("P"("L" ∩ "C"))/("P"("L"))`
= `("P"("C") * "P"("L"//"C"))/("P"("L"))`
`= (square * square)/square`
`= square`
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