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There are $15$ people from Swansea, $12$ from Wolverhampton and $10$ from Aberystwyth. A committee is to be made of at least $5$ people from Swansea and $2$ from anywhere else. The committee will only form if at least 5 people from Swansea are on it. What is the probability that the committee is entirely from Swansea (i.e. all $7$ people on the committee are from Swansea) ?

The way I worked it out was by writing out all of the possible combinations: $$^{15}C_7 +\left(^{15}C_6 \times ^{12}C_1\right) + \left(^{15}C_6 \times ^{10}C_1\right) + \left(^{15}C_5 \times ^{10}C_1 \times ^{12}C_1\right) + \left(^{15}C_5 \times ^{10}C_2\right) + \left(^{15}C_5 \times ^{12}C_2\right)~.$$ This equals $810,238$. $($Note: $^{15}C_7$ means $15$ choose $7~)$

I then did $^{15}C_7$ divided by $810,238$ and I got $0.79\%~$. I don’t understand why some people think you only need to focus on the last two positions, in which case you get 9% (as 10/32 x 9/31 = 45/496), as a committee would also form if the 2nd, 3rd, 4th, 5th and 6th people were from Swansea so you would only need to focus on the first and last positions. I know I must have gone wrong in my reasoning but if you could point out where and how I would be very grateful.

I was wondering if the method shown above is even the correct method and, if it is, if there is a quicker way of working it all out (if it’s correct).

AOD
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    It is far easier to group the wolverhampton and aberystwyth people together into a single category of "Non-swanseans." This will simplify your cases consideribly down into merely three cases. Either five people are from swansea and the other two are from elsewhere, six people are from swansea and one other from somewhere else, or all seven are from swansea. As for "why some people think you only need to focus on the last two positions" you are absolutely correct that this is a mistake to do and they are wrong for doing so. – JMoravitz Jan 29 '20 at 23:57

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You are assuming that 7 people are randomly selected and then the committee is checked for validity.
The "alternative" solution assumes that 5 people from Swansea are placed in the committee to assure the validity of the committee and after that the remaining 2 members are selected.
I find the wording of the problem vague enough to leave room for both interpretations (yours seems more "mathematical" - the alternative feels more realistic).

shortmanikos
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