6 ms·
Oddly enough that he should mention this, but last week or so I was thinking, "my how I have lost some of my mathematical prowess". There was a room of 4 males
by chegra 16y ago
Oddly enough that he should mention this, but last week or so I was thinking, "my how I have lost some of my mathematical prowess".
There was a room of 4 males and 10 females, now there was 5 other guys who were to be paired with someone in that room. What is the probability that the pairs would be as follows
(MM,MM,MM,MF,MF) where M=Males and F=Females.
This was a real life problem I create for myself to determine the bias by the selector. I struggled to work it out by pen and paper. This was a problem 8 years ago it wouldn't take a minute to do, most of the work would be in punching numbers in the calculator. Eventually, I said when I go home I will program it up. Between the time it took for a friend to restart her computer, I had whipped up a simulation. This is something I wouldn't have been able to do 8 years ago.
So in a sense, I call it a skills exchange. I lose some skills and gain some skills. I'm sure with practice I would be back up to par, the same goes for this guy.
- light3 16y agoIs the answer: 4x3x2x10x9 / (14x13x12x11x10) * 5! Is there some easier way to solve this?
- piaskal 16y agoIs there any difference between x and * operations?
- light3 16y agoNo, if I use * with no space it doesn't show.
- chalst 16y agoThe complexity of the formula seems to match the complexity of the problem: three lists of decreasing integers matching the three groups, multiplied/divided as appropriate. The hard bit is having the intuition to see that this is right, which is just what Bill Dubuque's answer goes on about in the linked article.
- malbertife 16y agoYou need to compute the ratio between the number of successful experiments and the number of total experiments. If what you want to know is the probability of having the 5 guys paired with 3 men and 2 women (which is what I think the grandfather meant), the answer is C(4,3) x C(10,2) / C(14,5) = about 9% (number of ways to pick 3 men out of 4 x number of ways to pick 2 women out of 10 / number of ways to pick 5 people out of 14, /regardless of the order/) If you want to know the probability of having the pairs made exactly in the order the grandfather mentioned, it's P(4,3) x P(10,2) / P(14,5) = about 0.9% (as above, but considering the order) where C(M,N) and P(M,N) are the number of combinations and permutations of N items out of M, respectively. (edit to change multiplication symbol, asterisk doesn't show)
- deleted 16y ago[deleted]
- ohmygodel 16y agoYou need to divide by 3! and 2!. One way to look at it is that every selection of five from the 4M and 10F is equally likely. There are \binom{14}{5} (i.e. 14 choose 5) such selections. Of these, \binom{4}{3} * \binom{10}{2} have exactly 3M and 2F. Therefore, the probability is \binom{4}{3} * \binom{10}{2} / \binom{14}{5}, or (4 * 3 * 2 * 10 * 9) 5! / (3! 2! (14 * 13 * 12 * 11 * 10)) = 10 * 9/(13 * 11 * 7) = 0.08991...
- dcosson 16y agoThis is how I'd solve it too. Plus, with a username like that and a LaTex formatted answer, I'm already inclined to trust your math skills :)
- light3 16y agoAhh indeed, I thought there were 5!=120 ways to order 3M and 2F, but apparently there are only 5!/(2!3!) = 10 ways: {MMMFF,MMFFM,MFFMM,FFMMM,FMMMF,MFMMF,MMFMF,FMMFM,MFMFM,FMFMM}. I made the mistake of thinking you can distinguish between individual males/females.