smallest number of socks to guarantee that the selection contains at least $10$ pairs A drawer in a darkened room contains $100$ red socks, $80$ green socks, $60$ blue socks and $40$ black socks. 
A youngster selects socks one at a time from the drawer but is unable to see the color of the socks drawn.
What is the smallest number of socks that must be selected to guarantee that the selection contains at lest $10$ pairs?
(A pairs of socks is two socks of the same color. No sock may be counted in more than one pair.)
What is the smallest number of socks that must be selected to guarantee that the selection contains at least $10$ pairs?
My attempt:
$100 = a $
$ 60 = b$
$ 40 = c$
$ 80 = d$
$a + b +c +d =280 $ socks
I know the probability of choosing each color on the first try are :
$p(a) = 0,3571;\,\,  p(b) = 0,2142;\,\, p(c) = 0,1428;\,\, p(d) = 0,2857$
How can I find  the smallest number of socks that must be selected to guarantee that the selection contains at lest $10$ pairs?  

Suppose you put $n$ socks into $4$ color boxes such that there are a total of exactly $k$ pairs of socks in the $4$ color boxes.  Then, $n$ must be at least $2k$ because that many socks are required for the $k$ pairs.  The maximum possible value of $n$ is $2k+4$ because each of the $4$ color boxes can have an odd number of socks and so there can be one unused socks in each color box.  Thus, $2k \le n \le 2k+4$.  If $k=9$, then $18 \le n \le 22$.  Thus, the maximum number of socks we can have in the $4$ color boxes and still not have guaranteed $10$ pairs is $22$.   So the answer to your question is $23$.
