In a multiple-choice test, each question has 5 options of which exactly one is correct. A student knows the answer to a question with probability \(\displaystyle\frac{3}{5}\); if he knows it, he marks the correct option, and otherwise he guesses one of the 5 options at random. If he has marked the correct option, then the probability that he knew the answer is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{15}{17}\)
Let \(E_1\) be the event that the student knows the answer, \(E_2\) the event that he guesses, and A the event that he marks the correct option.
\( \displaystyle P(E_1) = \frac{3}{5},\quad P(E_2) = \frac{2}{5} \)
\( \displaystyle P(A|E_1) = 1,\quad P(A|E_2) = \frac{1}{5} \)
By the theorem of total probability:
\( \displaystyle P(A) = \frac{3}{5}\times 1 + \frac{2}{5}\times\frac{1}{5} = \frac{15}{25}+\frac{2}{25} = \frac{17}{25} \)
By Bayes’ theorem:
\( \displaystyle P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(A)} = \frac{\frac{15}{25}}{\frac{17}{25}} = \frac{15}{17} \)
Why the other options are incorrect
Option (2) \(\displaystyle\frac{17}{25}\) — Incorrect. This is P(A), the overall chance of a correct answer, not the chance that he knew it.
Option (3) \(\displaystyle\frac{3}{5}\) — Incorrect. This is the chance of knowing before we see that the answer is correct; a correct answer makes knowing more likely.
Option (4) \(\displaystyle\frac{2}{17}\) — Incorrect. This is \(P(E_2|A)\), the chance that the correct answer was a lucky guess.