Combinatorial Probability

Hello, I am having trouble on one of my homework problems at my university. It involves binomial and multinomial coefficients. The problem goes as following:

Suppose that an airport food court has three restaurants: Subway, Burger King, and Mcdonalds.

if subway is closed and each customer is equally likely to go to either Mcdonalds or Burger King, what is the probability that six of the next nine customers go to Burger King and three of the nine go to Mcdonalds?

Right now I am stuck in how to approach the problem. I was thinking to start with 9*9 possibilities of each customer going to either restaurants. Am I on the right track? What else do I need to consider and think about?

Help is is greatly appreciated!

Re: Combinatorial Probability

Ok, let's break it down with the likelihood of where a single customer will go. Let P(S), P(B), P(M) be the probabilities that a single customer will go to Subway, Burger King, and McDonalds respectively. If Subway is closed, what is P(S)? We are given that P(B) = P(M) since a customer is equally likely to visit each. Assuming every customer visits exactly one establishment, we have P(S) + P(B) + P(M) = 1. Next, we have nine customers to observe. If we write down their choices of venues as a list with each choice an element of {S,B,M}, we obtain a list like this: S, S, B, M, M, B, M, B, M. The probability that the nine customers will go to the restaurants in that order is given by P(S)P(S)P(B)P(M)P(M)P(B)P(M)P(B)P(M) = $\displaystyle \left( P(S) \right)^2\left( P(B) \right)^3\left( P(M) \right)^4$. Now, if we were to take those same letters and rearrange their orders (but keep the same number of each letter), this would get the exact same probability. So, to determine the probability of 2 of the customers going to Subway, 3 of the customers going to Burger King, and 4 of them going to McDonalds, it is the probability we already calculated, but by the Sum principle, we will add that figure a number of times equal to the number of possible ways we can order the letters S,S,B,B,B,M,M,M,M.

Next, the question is asking for the probability that six of the nine customers go to Burger King and three of the nine go to McDonalds. So, out of the nine positions, we must choose six of them to be B. Then we have three positions remaining. Of those three positions, we must choose three of them to be M. Then, we multiply that by the probability of the first six go to Burger King and the last three go to McDonalds.

Re: Combinatorial Probability

Quote:

Originally Posted by

**Jaysquared** Suppose that an airport food court has three restaurants: Subway, Burger King, and Mcdonalds.

if subway is closed and each customer is equally likely to go to either Mcdonalds or Burger King, what is the probability that six of the next nine customers go to Burger King and three of the nine go to Mcdonalds?

This is a simple binomial probability. The probability is $\displaystyle (0.5)$ for either.

So $\displaystyle \binom{9}{6}(0.5)^9$