Math Is Fun Forum

MichaelWright

Guest

Tricky (im)probability question

Hi folks - I need some help with a tricky probability. Here's the situation:

Let's say there are 4M internet users in Age Group A. (The total set)
Of those 4M, there are 1,000 users who play a specific sport.
Those 1,000 are spread evenly over 125 teams, so 8 players each.

1. What's the probability of selecting 3 random users (from the 4M) who all play that sport?
2. What's the probability of all 3 users being members of the same team?

This isn't a school problem - I'm a 47 year old writer, and I need to figure out this absurdly improbable number. (The improbability is the point.) Any advice / guidance / direction here would be appreciated! (I have a moderate familiarity with combinations & permutations, and their notations, so do feel free to use whatever formulae / notations are needed. Ideally, a clear formula with values I can input would be great; a pointer to an online calculator that could crank out answers would be awesome for future uses, too.)

Any questions, please feel free to let me know. Thanks! ~ MW.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,140

Re: Tricky (im)probability question

hi Michael,

Welcome to the forum.

To choose one player from 4M is a probability of 1000/4000000.

To choose a second (one less to choose from, and one less in the grand set) is 999/3999999

And for a third, 998/3999998

Thus to end up with three chosen is (1000x999x998)/4000000 x 3999999 x 3999998)

But, lets say you do succeed with player A, then B then C.  You'll get the same combo if you choose B first, then A then C, so the same result has been calculated more than once.  So you need to also calculate how many ways you can re-arrange three players amongst themselves.  ABC, ACB, BAC, BCA, CAB, CBA or perm 3 which is 3!.

The second part is similar but taking account of the same team requirement.  There's no restriction when choosing the first as you haven't specified which team,  so 1000/4000000 again.

But the second player must also be in that team, so now there's only 7 to choose from. ie. 7/3999999.

And for the third, just 6 players to choose from so 6/3999998.

Again multiply together and divide by 6.

Hope that helps,

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob