Given that b<2k and g<2k, what is the probability that within the group of theatregoers there will be more girls than boys?
No Solution Yet  Submitted by Ady TZIDON 
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re(3): Solution or partial solution ... clarification 

Charlie,
Though your answer may be correct for the sample case, the solution, itself, is difficult to follow and is not what I would deem clarification. For example for 5 tickets, I see evidence of the use of factorials as used in combinatorics [g!/(g+b)! = (14/20)*(13/19)*(12/18)*...], but the numerator stops at 13 and the denominator skips 9,8,7,3, and 2. Apparently, some of the figures have been factored out, but nothing to explain why? Sometimes it would be better to just present an answer than an incomplete solution. Of course, looking at what had been presented before did hint that the 5 ticket portion of the solution was missing. Yet, without proper clarification, I felt unsure if any of the terms were simply missing (Thank you, Steve for also observing what I had seen). Though I understand the basic principles of combinatorial math, I am no expert, and I apologize that today and yesterday I have been too lazy to do the research to find what factors were involved.
Ady,
Due to the sample case not conforming to the givens, I am still unclear how one can approach a definitive answer for the general case. An assumption can be made that all tickets given to the class are used with no excess tickets [(b+g) = 2k], all tickets are used, but there are excess tickets [(b+g) < 2k], or the possibility of more students than available tickets [(b+g) > 2k]. As there is no limit to the number of possibly excess tickets (the value of k is not restricted), I am unsure whether a definitive answer can exist for a case where excess tickets were permitted, and perhaps it is to be inferred from such that the case was not meant for consideration.
Posted by Dej Mar on 20141010 03:19:43 