Combinatorial Probmlem

  • Thread starter HF08
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  • #1
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Hi,

I was reading a models book and read about a study where 9 analysts were chosen at random. Out of this group, 3 were selected for 2 week training, 3 were selected for 3 week training, and 3 were selected for 5 week training. Now I believe this models course had the idea that the first group woud be 9 choose 3. The second, 6 choose 3, and the last one 3 choose 3. I decided to ask myself a question.

This was a linear regression model type problem, but I couldn't help wondering how it would stink if you had to train for 2, 3, and 5 weeks. That is, a total of 10 weeks. So, I am going to ask the following:

Assume you could end up training for 2, 3, and 5 weeks. What is the chance you would be unlucky that they would make you study for 2 weeks, 3 weeks, and 5 weeks? What is the probability that you would be chosen for no weeks given this scenario? Also, what is a way I could say, you were chosen to train for 2 and 5 weeks, but not 3 weeks?

Thanks,
HF08
 

Answers and Replies

  • #2
CompuChip
Science Advisor
Homework Helper
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So the original problem is about combinatorics without repetitions, while the new one is a combination of no repetitions, and permutation with repetitions. It's like drawing balls from a vase: in the original problem, each successive draw is taken from the remaining balls; in your new problem, you put each ball back in the vase before picking the new one (which introduces a chance of picking the same ball again).

It's basically like three separate draws without repetition (apply the combinatorics of the original question) for each training course (2, 3 and 5 weeks).
 
  • #3
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So the original problem is about combinatorics without repetitions, while the new one is a combination of no repetitions, and permutation with repetitions. It's like drawing balls from a vase: in the original problem, each successive draw is taken from the remaining balls; in your new problem, you put each ball back in the vase before picking the new one (which introduces a chance of picking the same ball again).

It's basically like three separate draws without repetition (apply the combinatorics of the original question) for each training course (2, 3 and 5 weeks).


Let's work on the 2 weeks without replacement. There are C(9,3) ways to take out 3 people from a group of 9. This is where I get stuck. How can I calculate the odds of an individual being picked as 1 of 3 for the 2nd week?

I have difficulty with understanding combinatorics/permutations. So your help my own question as well as resources to read would be appreciated.

Thanks,
HF08
 

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