Math
posted by Genie .
A drawer has 10 pairs of gloves. If I grab 5 gloves at random, what is the probability that I pick at least one matched pair? What is the probability that I pick at least one right glove and one left glove?

let's look at the prob that they are all different
start by picking any glove, now you have 1
there is 1 of the remaining 9 that will match
we don't want that, so the prob that the 2nd is NOT a match is 8/9
prob that the 2nd and third are NOT a match
= (8/9)(7/8)
prob that the 2nd, 3rd, 4th and 5th are NOT a match
= (8/9)(7/8)(6/7)(5/5)
so the prob that at least one match is found
= 1  (8/9)(7/8)(6/7)(5/5)
= .....
Prob(at least one right and one left glove)
implies we don't want either all lefts or all rights
there are 5 lefts and 5 rights
prob(5 lefts) = (5/10)(4/9)(3/8)(2/7)(1/6)
the same would be true for prob(5 right)
so prob (all left or all right)
= (5/10)(4/9)(3/8)(2/7)(1/6) + (5/10)(4/9)(3/8)(2/7)(1/6)
= 2(5/10)(4/9)(3/8)(2/7)(1/6)
so prob(at least one left one right)
= 1  2(5/10)(4/9)(3/8)(2/7)(1/6)
= ... 
in the first solution near the end
= (8/9)(7/8)(6/7)(5/5)
so the prob that at least one match is found
= 1  (8/9)(7/8)(6/7)(5/5)
should have been:
= (8/9)(7/8)(6/7)(5/6)
so the prob that at least one match is found
= 1  (8/9)(7/8)(6/7)(5/6) 
Wouldnt it be from 20 gloves as we have 10 pairs.
So for first glove we can pick any. [19 remaining]
For second glove we have to select from 18 which are different  18/19
3rd glove  16/18
4th glove  14/17
5th glove  12/16 ... 
Of course you are right, how silly of me, there are obviously 20 gloves.
But.... why are you jumping from 18/19 to 16/18 etc
the pattern still continues following my argument above.
that is ...
(18/19)(17/18)(16/17) .... 
Ohk so it will be 
1st glove  20
2nd glove  18/19
3rd glove  17/18
4th glove  16/17
5th glove  15/16
Total probability = 1  the probability of above things right ??