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Question: Given $5$ people in an elevator on the ground floor, and the buttons for the second, third, and forth floors are lit, what is the probability that two people will exit the elevator at floor $3$?

Maths noob here. I have spent about $10$ minutes trying to come up with a solution to this question. I intuitively think the answer is $0.6$ but I lack the tools to reason about it properly. Any help? Apologies for the "please do my work for me" question.

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    $\begingroup$ *assuming nobody enters on the second floor, we have a uniform prior over the floor popularity, and there are no mistakenly-lit buttons $\endgroup$ – MichaelChirico Jan 19 '18 at 1:33
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In probability there is something called "sample space". It is, very simply, the set of all possible outcomes of a situation.

In this case all possible outcomes are these (assuming that at least one person exits at the floors that have a lit button):

$$ \begin{matrix} Floor 1 & Floor2 & Floor 3 \\ 1&1&3 \\ 1&3&1\\ 3&1&1 \\ 2&2&1 \\ 1&2&2 \\ 2&1&2 \\ \end{matrix} $$

In the first outcome $1$ person exits the elevetor at the first floor, $1$ person exits the elevator ar the second floor and $3$ people exit the elevator at the third floor. Total: $5$ people. The other outcomes are the other combinations.

If all the outcomes have all the same probability of happening, to calculate the probability we count in how many outcomes there are $2$ people exiting the elevator at the third floor. If you look at the table there are only $2$ such cases (the last ones, assuming the problem is asking for exactly two people, and not at least two people). Then we divide this number by the total number of outcomes.

So the probability that two people will exit the elevator at the third floor is: $$2/6 = 1/3$$

As you can see, when the outcomes have all the same probability of happening, a probability of an event is nothing else than the ratio between the number of outcomes of that event and the total number of outcomes.

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  • $\begingroup$ Thanks for that. Really clear. $\endgroup$ – tom redfern Jan 18 '18 at 11:45
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    $\begingroup$ Note that this is not the same as my interpretation. In this answer you assume that the pattern $(1,1,3)$ has the same probability as $(1,2,2)$. However, there are $20$ different ways to assign floors to people with a pattern of $(1,1,3)$, and $50$ different ways which give a pattern of $1,2,2$. $\endgroup$ – Especially Lime Jan 18 '18 at 11:51
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    $\begingroup$ @tomredfern The assumptions in the other answer are probably closer to the ones intended by the excercise, so I suggest you to read that carefully too and spot the differences. $\endgroup$ – emanuele ascani Jan 18 '18 at 13:29
  • $\begingroup$ Slightly nitpicky but it's a sample space not a space sample. $\endgroup$ – DRF Jan 18 '18 at 14:13
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    $\begingroup$ Just to have it clearly laid out for OP - this answer assumes that we don't care who wants what floor, we only care about the number of people who leave at each floor; treating the people as identical objects. @EspeciallyLime's answer assumes that the identity of each person leaving at each floor matters. Neither is right or wrong (though I would've assumed the question was looking for the first one), they are solutions to two different problems. Hopefully you can decipher which is the one your instructor/book/resource intended. $\endgroup$ – HammerN'Songs Jan 18 '18 at 18:04
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There isn't really enough information in the question. A reasonable way to fill in the gaps in a precise manner would be as follows.

Five people get into an elevator on the ground floor. Each person is equally likely to want to go to any other floor, independently of the other people. Given that at least one person wants each of floors $2$, $3$ and $4$, and no-one wants any other floor, what is the probability that exactly two people want floor $3$?

(It might be that the intended interpretation was "at least two" rather than "exactly two".)

We can put the people in order, and write down the sequence of floors they want to go to. A sequence might be $3,2,2,4,2$ or $2,3,4,2,3$ (order matters).

We have two questions to solve: how many possible sequences are there, and how many of these have exactly two $3$s? Then the second number divided by the first gives the probability, since all sequences are equally likely.

Hint for the first: there are $3^5$ sequences using only $2$s, $3$s and $4$s - try to count how many don't use each floor at least once, and subtract.

Hint for the second: there are $\binom 52$ ways to choose the two people who want floor $3$. For each choice, we then have to assign $2$ or $4$ to each of the other three, but we can't assign the same floor to all three of them.

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    $\begingroup$ Mostly OK, but what is the probability that one of the 5 occupants on the elevator punched a button by mistake? ;-) $\endgroup$ – MaxW Jan 19 '18 at 0:32

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