# Independence and Conditional Probability

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• Oct 13th 2013, 03:54 PM
BadWolf
Independence and Conditional Probability
I'm having difficulty answering this question:

Suppose you are one of 42 people attending a c
onference. At every meal, the conferenceattendees are seated randomly at 7 tables, with 6 people at each table (so that you sit withfive other people at each meal). What is the probability that, at the next three meals, youwill sit with 15 different people?

We know that there are 6^7 ways of seating 6 people in 7 tables but then I don't know where to go from there.
• Oct 13th 2013, 04:33 PM
Plato
Re: Independence and Conditional Probability
Quote:

Originally Posted by BadWolf
I'm having difficulty answering this question:
Suppose you are one of 42 people attending a conference. At every meal, the conferenceattendees are seated randomly at 7 tables, with 6 people at each table (so that you sit withfive other people at each meal). What is the probability that, at the next three meals, youwill sit with 15 different people? We know that there are 6^7 ways of seating 6 people in 7 tables but then I don't know where to go from there.

This is much more complicated than you can imagine.
There are $\frac{42!}{(7!)^6\cdot 6!}\text{ ways to place forty-two people at seven tables.}$

These are known as unordered partitions.

Now how do you count your being in totally different groups in three times?
• Oct 15th 2013, 07:37 PM
BadWolf
Re: Independence and Conditional Probability
(41 choose 5) ^3 ?