# Math

posted by on .

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?

• Math - ,

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)
= ...

• small typo - Math - ,

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)

• Math - ,

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 ...

• Math - ,

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) ....

• Math - ,

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 ??