Probability of teacher being in the last room: $\frac 45$ divided by $8$ A student is looking for his teacher. There is a 4/5 chance that the teacher is in one of 8 rooms, and he has no specific room preferences. Student checked 7 of the rooms, but the teacher wasn't in any of them. What's the probability that he is in one of the 8 rooms?
I tried dividing the P(4/5) by 8 and getting probability of teacher being in any one room of 0.1, and then subtracting 0.1*7 from 1 to get 0.3 - probability that he is in the last room. However that's not the right answer.
 A: You have found out:

*

*$0.1$ is the probability that the teacher is in a particular room, for any of the $8$ particular rooms (assuming that we haven't started looking for him).

*$0.7$ is the probability that the teacher is in one of the first $7$ rooms checked.

*$1-0.7$ is the probability that the teacher is not in one of those rooms i.e. that he is in the last room or not in any of those rooms.

What you are actually asked for is the probability that the teacher is in the $8$th room, given that the first $7$ rooms are empty.
Before we checked the rooms, there was a $0.1$ chance that the teacher was in the $8$th room, and a $0.2$ chance that the teacher was not in the $8$ rooms.
After we checked the rooms, one of these two options must be the case, and the ratio of their likelihoods remains the same, so the chance that it is the $8$th room is:
$$\frac{0.1}{0.1+0.2}=\frac{1}{3}$$
The chance that the teacher is not there is $2/3$, preserving that this option is twice as likely as the teacher being in a particular room.
In general, this idea of relative likelihood being preserved is encapsulated in Bayes' theorem, which you could use to formulate the same answer.
A: Hint:
Number the rooms with $i=1,2,\dots,8$.
Let $E_i$ denote the event that the teacher is in room $i$.
Now find:$$P(E_8|E_1^c\cap\cdots\cap E_7^c)$$
Beware that the events are mutually exclusive with:$$P(E_1\cup\cdots\cup E_8)=\frac45\text{ and }P(E_1)=\cdots=P(E_8)$$
