# Can I apply Bayes' Theorem here?

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?

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