Probability: binomial coefficient problem

[Jaysquared]

Hello physics forum! I come to you with a binomial coefficient problem I am stuck on. Here is the question

1. Suppose an airport has three restaurants open, Subway Burger King and McDonalds. If all three restaurants are open and each customer is equally likely to go to each one, what is the probability that out of 9 customers, three will go to each one?


2. Permutations and binomial coefficient comes in handy for this



3. Alright so my initial idea is that since all three restaurants are open, there is a 1/3 chance of choosing one of them. Since there are 9 people choosing one of the three restaurants, It would be (1/3)^9. Using the binomial coefficient the initial setup would be 9 choose 3. Am I on the right track? The answer that they give is 0.085. Please help for any suggestions. Thank you!

 

[jbunniii]

Let's introduce the following notation:
$A = \{$ exactly 3 people dine at Subway $\}$
$B = \{$ exactly 3 people dine at BK $\}$
$C = \{$ exactly 3 people dine at McD $\}$
Then you are trying to calculate $P(A \cap B \cap C)$. This is the same as $P(A \cap B)$ (why?). Try calculating it using $P(A \cap B) = P(A | B)P(B)$.

 

[Ray Vickson]

Let's introduce the following notation:
$A = \{$ exactly 3 people dine at Subway $\}$
$B = \{$ exactly 3 people dine at BK $\}$
$C = \{$ exactly 3 people dine at McD $\}$
Then you are trying to calculate $P(A \cap B \cap C)$. This is the same as $P(A \cap B)$ (why?). Try calculating it using $P(A \cap B) = P(A | B)P(B)$.

Alternatively: use the multinomial distribution (trinomial in this case); see, eg., http://en.wikipedia.org/wiki/Multinomial_distribution or http://mathworld.wolfram.com/MultinomialDistribution.html