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?
Best Answer
Let the different types of socks be represented by $\text{A},\text{B},\text{C}$ and $\text{D}$
Suppose that you wish to draw one pair of socks from the drawer. Then you would pick $5$ socks (one of each kind, plus one (Let $\text{A}$) to guarantee atleast one pair).
Notice that in the worst possible situation, you will continue to draw the same sock $A$
$\big($Because as you draw any other sock, let $\text{B}$, it will combine from previously selected $\text{B}$ sock and will result in a pair just by adding one sock, while in the case if selecting sock $\text{A}$ you'll have to select a total $2$ sock after those $5$ to make a pair$\big)$ , until you get $10$ pairs. This is because drawing the same sock results in a pair every $2$ of that sock, whereas drawing another sock creates another pair. Thus the answer is