A family consisting of four persons—A, B, C, and D—belongs to a medical clinic that always has a doctor at each of stations 1, 2, and 3. During a certain week, each member of the family visits the clinic once and is assigned at random to a station. The experiment consists of recording the station number for each member. Suppose that any incoming individual is equally likely to be assigned to any of the three stations irrespective of where other individuals have been assigned.

What is the probability that one station has two family members at it and the others have only one?

Answer: .444

I have the answers and see multiple ways to get to the answer, but it might just be a coincidence or some relation I'm not seeing. I've tried binomial method but it wasn't consistent to get other answers. I just need a clear method for these type of problems.

  • 1
    $\begingroup$ There are two valid ways to interpret the question, with two different answers. (1) Given a particular station (viz., the "one station") what is the chance that two members are assigned to it and the other two stations get one member each? (2) What is the chance that there exists some station with two members assigned to it (and the other two stations get one member each)? Because there are three equivalent stations, the answer to (2) must be three times the answer to (1). So, before attempting an answer, you first need a clear method to state this type of question! $\endgroup$ – whuber Aug 19 '14 at 21:47

This seems to be a question using the multinomial distribution. In essence you have 4 trials (the family members) to be split into 3 categories (A, B, C). Each category has an equal chance of success (i.e. the family member being assigned there). Let the number of members assigned to A, B and C be $n_A$, $n_B$ and $n_C$ respectively. Then the relevant probability mass function is:

$$f(n_A,n_B,n_C) = \frac{4!}{n_A!n_B!n_C!}\left(\frac{1}{3}\right)^4$$

You want two members assigned to one station, and the other two assigned to separate stations. This can be done in three ways, so the required probability is:

$$3\times \frac{4!}{2!1!1!}\times \left(\frac{1}{3}\right)^4 = \frac{4}{9} \approx 0.444$$


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