This seems like a easy probability question, yet I can't seem to wrap my head around it. I don't know if the total amount of tries is $5!$ or if there is a total number for each different case.
Ms. Smith had $5$ keys on her key chain. There are $n$ key(s) to open her classroom. But, Ms. Smith forgot which one is the right one, so she tried one by one.
- if $n=1$, what is the probability that Ms. Smith can open the door at the third try?
- if $n=1$, what is the probability that Ms. Smith can open the door within three tries?
- if $n=2$, what is the probability that Ms. Smith can open the door at the third try?
This is selection without replacement.
You have $n$ opening keys $O$ and $5-n$ non opening keys $N$.
a) The possiblities are $O, NO, NNO, NNNO,NNNNO$ with respective probabilities ${1 \over 5}, {4 \over 5}{1 \over 4}, {4 \over 5}{3 \over 4}{1 \over 3}, {4 \over 5}{3 \over 4}{2 \over 3}{1 \over 2}, {4 \over 5}{3 \over 4}{2 \over 3}{1 \over 2}$. By inspection see see that the probabilities sum to one.
In any event, the probability you want is ${4 \over 5}{3 \over 4}{1 \over 3}$.
b) This is just ${1 \over 5}+ {4 \over 5}{1 \over 4}+ {4 \over 5}{3 \over 4}{1 \over 3}$.
c) The possibilities are different now, and they are $O,NO,NNO,NNNO$, with respective probabilities ${2 \over 5}, {3 \over 5}{1 \over 2}, {3 \over 5}{1 \over 2}{2 \over 3}, {3 \over 5}{1 \over 2}{1 \over 3}$. Again, by inspection see see that the probabilities sum to one.
The answer is the third probability ${3 \over 5}{1 \over 2}{2 \over 3}$.