Odds of group getting chosen

[Tack]

Eleven groups are due to present their project in History class. Three groups are chosen to present each day. If the teacher chooses groups to present at random, what are the odds in favor of your group having to present on the first day?

 

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[Tack]

Starting with a small example. 4 groups (A B C D) and choosing 2 groups give the following permutations:
AB BA AC CA AD DA BC CB BD DB CD DC. or 12 possible permutations of 2 items. Since A appears 6 times, the probability of group A being chosen is 6/12. The denominator is easily found by (4P2). Since A being chosen has a probability of 3/12 for any slot(3P1/4P2), summing both slots gives 6/12.
Next, using the general formula of k * [(N-1)P(k-1)/NPk), where N is the number of total groups and k is the number of groups chosen, then N=11 and k =3 giving a probability of 3/11. I derived this formula myself so not sure if it's correct. Unfortunately the answer given in the online example is 3/8. Any help is appreciated.

 

[Dr.Peterson]

Eleven groups are due to present their project in History class. Three groups are chosen to present each day. If the teacher chooses groups to present at random, what are the odds in favor of your group having to present on the first day?

Starting with a small example. 4 groups (A B C D) and choosing 2 groups give the following permutations:
AB BA AC CA AD DA BC CB BD DB CD DC. or 12 possible permutations of 2 items. Since A appears 6 times, the probability of group A being chosen is 6/12. The denominator is easily found by (4P2). Since A being chosen has a probability of 3/12 for any slot(3P1/4P2), summing both slots gives 6/12.
Next, using the general formula of k * [(N-1)P(k-1)/NPk), where N is the number of total groups and k is the number of groups chosen, then N=11 and k =3 giving a probability of 3/11. I derived this formula myself so not sure if it's correct. Unfortunately the answer given in the online example is 3/8. Any help is appreciated.

First, the answer is not the fraction 3/8 (the proper form for probability), but the ratio 3:8 (the proper form of odds); though some people do write odds that way, which can lead to confusion.

The naive approach quickly leads to this answer: There are 3 groups that are chosen, compared to 11-3=8 that are not.

I held off on replying yesterday, waiting to see if perhaps you were trying a harder way, and you are.

But your answer is in fact correct, as a probability! Just convert P = 3/11 to odds in favor = P : (1-P) = (3/11) : (8/11) = 3:8.

 

[Tack]

First, the answer is not the fraction 3/8 (the proper form for probability), but the ratio 3:8 (the proper form of odds); though some people do write odds that way, which can lead to confusion.

The naive approach quickly leads to this answer: There are 3 groups that are chosen, compared to 11-3=8 that are not.

I held off on replying yesterday, waiting to see if perhaps you were trying a harder way, and you are.

But your answer is in fact correct, as a probability! Just convert P = 3/11 to odds in favor = P : (1-P) = (3/11) : (8/11) = 3:8.

Thanks @Dr. Peterson for the correction. I learned 2 new things today. The help is much appreciated.