SOLUTION: I was wondering if I can enlist your help in a probability question. I should clarify that I am not a student. I think that my question may be expressed as: What is the probability

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Question 852789: I was wondering if I can enlist your help in a probability question. I should clarify that I am not a student. I think that my question may be expressed as: What is the probability to select four specific persons in terms of forming a committee of six persons from a list of 31 persons? I would think that the answer might be expressed as a fraction with its denominator being the number of all possible six-person committees that may be selected from the group of 31 persons and the numerator being the number of all possible couples that may be formed from the remaining 27 persons (excluding the four ones mentioned before). If this is correct, I estimate the probability as (27!) / (31!) and it seems to me that it is extremely small.
Found 2 solutions by stanbon, ewatrrr:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
What is the probability to select four specific persons in terms of forming a committee of six persons from a list of 31 persons?
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# of ways to select 4 specific persons:: 1
# of ways to select 2 addition persons:: 27C2 = 351
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P(condition) = 1/351
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Cheers,
Stan H.
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Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!
 

Hi,

'Combinations' are traditionally used for Your purposes of finding 

P(choosing 4 specific people plus  2 others from group of 31)

For ex: 31C6 represents total ways of choosing 6 people

 31C6 =  =   is equal to 

P(choosing 4 specific people plus  2 others from group of 31) = 

P =   = 351/736281 

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