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There are five hotels in a certain town. If 3 people check into hotels in a day, what is the probability that they each check into a different hotel?

Attempt

I was thinking to translate this problem into balls and boxes. So, we have $5$ boxes and $3$ balls that we put into this boxes. There are ${5 \choose 3}$ ways to put them balls into the boxes. Now, if the first ball is chosen first to go to any of the boxes then there are only $5$ ways to do it. And the second ball cannot go into the box chosen by the first one so it has only 4 ways to go into boxes and the third one must have just 3 ways to get into the boxes. Thus, (translating back into problem)

$$ P( \text{3 people different hotel} ) = \frac{5 \times 4 \times 3 }{ {5 \choose 3} }$$

Is this correct?

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    $\begingroup$ ${5 \choose 3} = \frac {5!}{3!2} = \frac {5*4}2 = 10$ so $\frac {5\times 4\times 3}{5\choose 3} = \frac {5\times 4 \times 3}{10} = 6$. So you have calculated a probability of $6 > 1$. So... no, that is not correct. $\endgroup$ – fleablood Apr 17 '18 at 17:50
  • $\begingroup$ "There are $5\choose 3$ ways to put them balls into the boxes" Really? $5 \choose 3$ assumes they are all in different hotels. If you are going to assume they must be in different hotels from the start then the probability is going to be 100 percent. $\endgroup$ – fleablood Apr 17 '18 at 17:53
  • $\begingroup$ Im very confused. maybe i need to think the size of sample space but istn the sample space just all possible selection of 3 choices of people for different hotels ? so that ${5 \choose 3}$ is how we count them $\endgroup$ – James Apr 17 '18 at 17:55
  • $\begingroup$ "Now, if the first ball is chosen first to go to any of the boxes then there are only 5 ways to do it." If there are $5 \choose 3$ ways to do it, then there are $5 \choose 3$ ways to do it and you have already figured it out. Why are you figuring it out a second time? If you think about it, what you are trying ot do simply makes zero logistic sense. What you want to do if figure out the number of ways for them to go into different hotels and divide by the number of ways they can go into hotels where maybe they go to the same hotel. $\endgroup$ – fleablood Apr 17 '18 at 17:56
  • $\begingroup$ "but istn the sample space just all possible selection of 3 choices of people for different hotels " No. The sample space it the ways any of the 3 people can go to any of the hotels. No need for the hotels to be different. The hotels being different are the outcome subset you are attempted to calculate. Your sample space is all possible outcomes. Not just the favorable ones. If you restrict to just the favorable ones you will get a probability 1 or higher. $\endgroup$ – fleablood Apr 17 '18 at 17:59
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The first person will check-in at a different hotel with a probability of 1.

The second person will check-in at a different hotel with a probability of 4/5.

The third person will check-in at a different hotel with a probability of 3/5.

So the probability that all check-in at different hotels are

$P = 1 * 4/5 * 3/5 = 12/25$

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(heavily edited) no thats wrong because you have taken the wrong sample space. the sample space would be 125 since each of them have 5 ways to choose a hotel.

so the final answer would be (5!/(2! 3!))*3!/125= 0.08*6 = 0.48

the event would be 5!/(2! 3!) since we are choosing 3 hotels out of five and we are multiplying 3! because the three guys can choose between three.

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  • $\begingroup$ do i need to edit the answer more? $\endgroup$ – Avnish Singh Apr 17 '18 at 18:14
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    $\begingroup$ Your answer is incorrect. You correctly calculated the size of the sample space. However, it matters which person chooses which hotel, so the number of favorable cases is not $\binom{5}{3}$. $\endgroup$ – N. F. Taussig Apr 17 '18 at 18:15
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    $\begingroup$ This tutorial explains how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Apr 17 '18 at 18:17

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