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I'm quite inexperienced in mathematics so I apologize in advance for using incorrect terms or fundamentally misunderstanding the application of Bayes' Theorem.

My question is about whether you can apply Bayes' Theorem in the given scenario:

  • A student is attempting to answer a question.
  • The underlying probability that a student will get a question in this category wrong is $50\%$. This is based on past data of how all questions have been answered so far.
  • A teacher thinks that the probability a student will get one specific question wrong is $70\%$. The teacher is accurate $90\%$ of the time.
  • What is the overall probability the student will get this specific question wrong?

My current understanding is that Bayes' Theorem gives this relationship:

$$ \text{Q wrong} = \text{"Student will get this specific question wrong"} \\ \text{T's guess} = \text{"Teacher thinks they will get this question wrong"} \\ $$

$$ P(\text{Q wrong}|\text{T's guess}) = \frac{P(\text{T's guess}|\text{Q is wrong})P(\text{Q is wrong})}{P(\text{T's guess})} $$

The interpretation being:

  • $P(\text{Q wrong})$ = Prior belief or "base rate"
  • $P(\text{Q wrong}|\text{T's guess})$ = Posterior, updated belief based on teacher's estimate

However, this model doesn't seem right. It doesn't fit the normal examples of Bayes' Theorem I've seen, like a medical test with a false positive rate since the teacher isn't just stating that they will get the question wrong, instead giving an estimate of the probability.

Very un-mathematically blindly plugging the numbers in doesn't work unfortunately:

$$ \frac{0.7 \times 0.5}{0.9} \approx 0.389 $$

Which isn't correct because the probability should be above the baseline estimate.

Is it possible to answer this question using Bayes' Theorem or using some other method? Am I missing something crucial that makes this question impossible to ask without other information?

Thanks in advance.

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    $\begingroup$ What does it mean for the teacher to be 90% accurate when he thinks that the student will "get one specific question wrong with probability 70%"? $\endgroup$
    – Benjamin Wang
    Feb 27, 2021 at 22:11
  • $\begingroup$ @BenjaminWang This is the kind of thing I didn't think of... thank you :) Maybe if it was said as "The teacher has predicted the probabilities of groups of students getting the question correct with 70% accuracy in the past"? $\endgroup$
    – Olly Britton
    Feb 27, 2021 at 22:19

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You actually need the law of total probability here. The sample space is partitioned into when the teacher is right and when the teacher is wrong. If the teacher is right, the student should get this specific question wrong with chance 70%. If the teacher is wrong, with probability 10%, we use the baseline probability that the student gets the question wrong (for there is no other estimate). We write

$$\begin{split}P(question\_wrong)&=P(question\_wrong|teacher\_right)P(teacher\_right)+P(question\_wrong|teacher\_wrong)P(teacher\_wrong)\\ &=.7(.9)+.5(.1)\\ &=.68\end{split}$$

This is between the baseline estimate and the teacher's shenanigans.

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    $\begingroup$ I think this is what I was looking for, it's kind of like giving weight to the different outcomes based on the teacher's accuracy. It makes a lot more sense like this, thank you! $\endgroup$
    – Olly Britton
    Feb 27, 2021 at 22:24

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