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14 Jun 2020, 04:46
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Dear IanStewart,
I meant using Combination (choosing a subgroup from a larger group) formula as quoted below:
BTW, I know that permutation works for this problem. But not sure about the combination one.
I meant using Combination (choosing a subgroup from a larger group) formula as quoted below:
Bunuel
One can also do: \(P=\frac{C^1_1*C^2_7}{C^3_8}=\frac{3}{8}\), where \(C^1_1\) is # of ways to select Kim, \(C^2_7\) is # of ways to select any 2 students out of 7 left and \(C^3_8\) is total # of ways to select 3 students from 8;
Or: you can find the probability that among 3 students selected to complete the tasks there won't be Kim and subtract it from 1:
\(P=1-\frac{C^3_7}{C^3_8}=\frac{3}{8}\);
I tried and failed to apply the above method to this question Or: you can find the probability that among 3 students selected to complete the tasks there won't be Kim and subtract it from 1:
\(P=1-\frac{C^3_7}{C^3_8}=\frac{3}{8}\);
BTW, I know that permutation works for this problem. But not sure about the combination one.
Updated on: 14 Jun 2020, 05:36
Originally posted by IanStewart on 14 Jun 2020, 05:14.
Last edited by IanStewart on 14 Jun 2020, 05:36, edited 1 time in total.
Last edited by IanStewart on 14 Jun 2020, 05:36, edited 1 time in total.
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varotkorn
Dear IanStewart,
I meant using Combination (choosing a subgroup from a larger group) formula as quoted below:
I meant using Combination (choosing a subgroup from a larger group) formula as quoted below:
Bunuel
1. One can also do: \(P=\frac{C^1_1*C^2_7}{C^3_8}=\frac{3}{8}\), where \(C^1_1\) is # of ways to select Kim, \(C^2_7\) is # of ways to select any 2 students out of 7 left and \(C^3_8\) is total # of ways to select 3 students from 8;
2. Or: you can find the probability that among 3 students selected to complete the tasks there won't be Kim and subtract it from 1:
\(P=1-\frac{C^3_7}{C^3_8}=\frac{3}{8}\);
I tried and failed to apply the above method to this question 2. Or: you can find the probability that among 3 students selected to complete the tasks there won't be Kim and subtract it from 1:
\(P=1-\frac{C^3_7}{C^3_8}=\frac{3}{8}\);
There are two different methods in that quote (I've numbered them in red). I wouldn't recommend using either for this type of problem -- it's easier just to think about making selections in order and multiplying, when possible. But you can use the above methods for the problem at the start of this thread -- edit: I accidentally answered the question "what is the probability Harry is chosen as President, Secretary or Treasurer?" in this post - the answer to that question is 3/10. I'll leave this solution up, but I answered the question from the OP, "what is the probability Harry is chosen as Secretary or Treasurer?" in a followup post below
1.: we have 10C3 possible choices in total for the first three people. If Harry will be one of our selections, we have 1C1 ways to choose him and 9C2 ways to choose the other two people, so the answer is 1C1*9C2/10C3 = 3/10
2.: we have 9C3 ways to choose 3 people different from Harry, and 10C3 selections possible in total, so the probability we don't pick Harry is 9C3/10C3 = 7/10, and the probability we do pick Harry is thus 1 - 7/10 = 3/10
14 Jun 2020, 05:34
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IanStewart
There are two different methods in that quote (I've numbered them in red). I wouldn't recommend using either for this type of problem -- it's easier just to think about making selections in order and multiplying, when possible. But you can use the above methods for the problem at the start of this thread:
1.: we have 10C3 possible choices in total for the first three people. If Harry will be one of our selections, we have 1C1 ways to choose him and 9C2 ways to choose the other two people, so the answer is 1C1*9C2/10C3 = 3/10
2.: we have 9C3 ways to choose 3 people different from Harry, and 10C3 selections possible in total, so the probability we don't pick Harry is 9C3/10C3 = 7/10, and the probability we do pick Harry is thus 1 - 7/10 = 3/10
1.: we have 10C3 possible choices in total for the first three people. If Harry will be one of our selections, we have 1C1 ways to choose him and 9C2 ways to choose the other two people, so the answer is 1C1*9C2/10C3 = 3/10
2.: we have 9C3 ways to choose 3 people different from Harry, and 10C3 selections possible in total, so the probability we don't pick Harry is 9C3/10C3 = 7/10, and the probability we do pick Harry is thus 1 - 7/10 = 3/10
Thanks to a PM from varotkorn, I realize I was answering the problem asked in the linked thread (but using the numbers in this one) rather than the question asked here (I got confused with so many tabs open linking to different questions!). I'll leave the above post up, but edit it to indicate what question I was answering. If instead you're answering the question asked in this thread, then you can still use the same methods, but we only care about the second and third selections, not the first, so the numbers get smaller. We have
1.: 10C2 ways to pick the Secretary and Treasurer, and 9C1*1C1 to make one of those selections Harry, so 9C1/10C2 = 1/5 is the answer
2.: 9C2 ways to pick the Secretary and Treasurer if Harry will not be selected, and 10C2 selections possible in total, so 1 - (9C2/10C2) = 1 - 4/5 = 1/5 is the answer
