The student answered correctly, did they actually know it?

Question

On a multiple-choice question with four options, a student either truly knows the answer (probability $0.6$ ) and answers correctly for sure, or does not know (probability $0.4$ ) and then guesses, getting it right with probability $1/4$ . The student answered correctly.

What is the probability the student actually knew the answer?

Accepted Answer

Split the "correct" outcome into its two possible causes. - Knew it and correct: 0.6 	imes 1 = 0.6. - Guessed and correct: 0.4 	imes 	frac14 = 0.1. The total chance of a correct answer is 0.6 + 0.1 = 0.7. Of that, the knowing case accounts for 0.6: P(	ext{knew} \mid 	ext{correct}) = \frac{0.6}{0.7} = \frac{6}{7} \approx 0.857. A correct answer is much more likely to come from real knowledge than from a one-in-four guess, so belief rises from the 60\% prior to about 86\%. Takeaway. A correct answer is evidence for knowledge, but a nonzero guessing chance means it is never full proof; Baye…

The student answered correctly, did they actually know it?

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