I want to solve the following:
During exam time, in a certain school, only 25% of the teachers warn their students in writing that they are not allowed to stand up and ask questions during the test. However, it has been observed that despite this warning, 20% of students do so. For mentors who do not provide such a warning, the corresponding figure is 70%. If during a test given by teacher X, an inspector bursts into the room and observes that there are students breaking the rule, what is the probability that the teacher did not warn in writing that it is forbidden to ask questions in the exams?
I asked about this exercise a few days ago, the link to that post is this one
In short, the answer was:
Let $A$ be the event that a teacher warns their students in writing about the rule. Let $B$ be the event that students get up to ask questions during the exam. We are given:
- $P(A) = 0.25$, which is the probability that a teacher gives a written warning.
- $P(A^c) = 0.75$, which is the probability that a teacher does not give a written warning
- $P(B|A) = 0.20$, which is the probability that students get up to ask questions given that they were warned.
- $P(B|A^c) = 0.70$, which is the probability that students get up to ask questions given that they were not warned.
We want to find $P(A^c|B)$, the probability that a teacher did not warn in writing, given that students are observed breaking the rule.
Bayes' Theorem is given by:
$$P(A^c|B) = \frac{P(B|A^c) \cdot P(A^c)}{P(B)}$$
To find $P(B)$, we need to calculate the total probability of students getting up to ask questions, which is:
$$P(B) = P(B|A) \cdot P(A) + P(B|A^c) \cdot P(A^c)= 0.20 \times 0.25 + 0.70 \times 0.75$$
So, we obtained that $P(A^c \mid B) \approx 0.538 $
BUT I came across the same exercise in a book, and they assume that the answer is $0.9130$ My question is, who is wrong, the book or me? I appreciate your help
