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I have question about probability. You have $3$ people in a elevator and a building with $6$ floors (floor 1-6). Each person leave the elevator in a random floor and they leave independently from other persons. What is probability:

  1. That all $3$ people leave elevator in same floor

  2. That in any floor exactly $2$ persons leave elevator

I have idea for 1. You have floor $6$ floors and $3$ people. You have how many possibles that people leave? You have $6^3$ different possible. Now you want all people in same floor, you have possible of $\frac{6}{216}$ because you have 6 floor and 216 different possibles. Is solution good?

  1. Is confuse me. Maybe is better I imagine this all like cube you throw $3$ times? And now you want throw any number $2$ times. For example you throw cube 1,1,2 or 6,4,4 or 5,5,5 etc. I need count all of these possibles and divide with 216 again. But don't know how? Is this correct how I believe?

Please help I write test next week...

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    $\begingroup$ Your solutions are correct, except 555 must not be counted because you want EXACTLY TWO people to leave at the same floor. Counting all possibilities for case 2. is not so difficult. $\endgroup$
    – Intelligenti pauca
    Commented Oct 16, 2017 at 17:05
  • $\begingroup$ @Aretino Thank you!! How I count it? I write all and count duplicate number? $\endgroup$
    – eyesima
    Commented Oct 16, 2017 at 17:07
  • $\begingroup$ @Aretino ahhh yes I readed wrong thanks for info! $\endgroup$
    – eyesima
    Commented Oct 16, 2017 at 17:08
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    $\begingroup$ the possibilities are 112,113,...221,223,...331,332,...6*5 in total. For each you can also rearrange in 3 ways. Thus 30*3/216=5/12. Or one minus P(all same) minus P(all different)=1-1/36-20/36=5/12 $\endgroup$
    – Xiangyu Chen
    Commented Oct 16, 2017 at 17:23
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    $\begingroup$ For the floor that you have 2 people on, it can be any from 1-6. Since the 3rd person cannot be on same, only 5 choices left for them. $\endgroup$
    – Xiangyu Chen
    Commented Oct 16, 2017 at 17:27

1 Answer 1

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You handled case 1 correctly

Of course also in case 2 you can count and divide, but in this answer I would like to promote another route.

If $E_i$ is the event that on floor $i$ exactly two persons leave the elevator then you are looking for $P(\bigcup_{i=1}^6E_i)$.

Now observe that the $E_i$ are mutually exclusive so that: $$P(\bigcup_{i=1}^6E_i)=\sum_{i=1}^6P(E_i)$$

Now realize that the $E_i$ are equiprobable so that: $$P(\bigcup_{i=1}^6E_i)=6P(E_1)$$

Finally realize that at floor 1 there are $3$ independent experiments that can succeed (i.e. person leaves elevator) with probability $\frac16$ or fail. Then we are dealing with binomial distribution with parameters $n=3$ and $p=\frac16$.

Final result takes the form:$$P(\bigcup_{i=1}^6E_i)=6P(E_1)=6\binom32\left(\frac16\right)^2\left(\frac56\right)^1$$

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