A person has 5 keys and they are trying to unlock a door where only 1 of the 5 keys will unlock the door. They choose a key at random, and if the key they choose does not unlock the door, they put that key away and choose another key at random from the remaining keys. They do this until they unlock the door. If X denotes the number of keys used to unlock the door, what is the probability distribution of X?
So, in my mind, $P(X = 1) = 1/5$, $P(X = 2) = 1/4$, so forth and so on. But, these probabilities do not add up to 1, so I must be wrong! This question then screamed at me that it may be hypergeometric, but when I try to compute $P(X = 2,3,4,5)$ I run into a problem where I need to calculate an $n \choose r$ where $r > n.$ So I am assuming my first intuition is incorrect. Then I thought, hey, this thing could possibly be binomial with $n = 5$ and $p = 1/5$. But then, $P(X = 1) > P(X=2)$ and I was like...nope this doesn't make sense because the probability of having to use more than 1 key should be bigger than having to use just 1. Then I realized I was being dumb because these events are totally not independent of each other. My last thought was that maybe - just maybe - this thing is a uniform distribution with $p = 1/5$ and I was using conditional probabilities to find what $P(X = x)$ was up above. Is this last ditch effort of mine correct? Any help is appreciated!!!
The sample space is so small that you can evaluate each probability directly:
$P(X=1)=\frac15=\frac15\\ P(X=2)=\frac45×\frac14=\frac15\\ P(X=3)=\frac45×\frac34×\frac13=\frac15\\ P(X=4)=\frac45×\frac34×\frac23×\frac12=\frac15\\ P(X=5)=\frac45×\frac34×\frac23×\frac12×\frac11=\frac15$
So $X$ does follow a uniform distribution. We do not need the calculations above, though: we can assume without loss of generality that the sequence of keys tried is fixed, so the problem becomes a linear search of the keys, and it is clear that the probability that the $n$th key tried opens the door ($P(X=n)$) is always $\frac15$.