There are 6 different rooms and 3 persons. Every person chooses a random room blindfolded. a) What is the probability that all of them go to different rooms? b) what is the probability that they go to the same room?

a) Is it wrong to say that there is a total number of 6*5*4 = 120 ways that this can happen? person 1 has 6 choices, person 2 has 5 choices, person 3 4 choices. Or rather 6^3 = 216?

b) Would it be right to say since there are 6 rooms, so there is only 6 different ways to pick. so P = 6/216?

I am a bit confused here. Can somebody clear this up please and the above question?


  • 1
    $\begingroup$ $6^3$ is the total number of ways that the people go to rooms. 120 in (a) is correct; (b) is correct too. $\endgroup$ – Parcly Taxel Sep 19 '17 at 17:44
  • $\begingroup$ To be a stickler, it should be emphasized in the problem statement that the choices are uniformly at random and independent. Without this, weird things can happen. For example, if our three persons are all old and get tired quickly and can't walk far, even though they are choosing rooms "randomly" they will prefer moving to one of the closer rooms. $\endgroup$ – JMoravitz Sep 19 '17 at 17:51
  • $\begingroup$ hi thanks for the responses. how would I go about if the question was changed to : what is the probability that 2 go in the same room and one in a different? What is the correct way to compute the number of ways? I might be wrong but I am just throwing a guess that it is 6C2 * 5C1? $\endgroup$ – Kev Sep 19 '17 at 19:11
  • $\begingroup$ No, your guess is incorrect. Put meaning behind each term you add or multiply together and try to phrase how you can describe each outcome that you are trying to count. $\binom{6}{2}\binom{5}{1}$ sounds like "picking two of the six rooms and then picking one of five rooms" which doesn't really make any sense. A correct setup: "Pick which room had two people going to it. Then pick which room had one person going to it. Finally, pick which of the three people was the one to go to the room by himself" $\endgroup$ – JMoravitz Sep 19 '17 at 19:35
  • $\begingroup$ Do you mind posting the solution to this? I honestly cannot work out a logical method/working to compute this..... (Person 1 - 6 ways to choose a room, *person 2 - 1 way, which is to go into the same room(?) and person 3 has 5 ways?) $\endgroup$ – Kev Sep 19 '17 at 20:58

Probability here can be found by finding the number of different ways your event can occur and dividing by total number of possible outcomes, which is $6^3=216$.

You are right, for (a), the number of ways this can happen is $6 \times 5 \times 4$ -- now compute the relevant probability.

For (b), there is obviously exactly 6 ways this can happen. Can you compute the probability here as well?


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