The answer is 69.4%. This is basically a counting problem in which we have to answer two questions: “How many possible combinations of 5 pairs of socks are there?” and “How many combinations include one of each color?”

We’ll start with the first question. She has 10 pairs and she’s choosing 5. Because the order in which she pulls the socks out of the drawer doesn’t matter, this is a combination written as C(10,5) and read “combination: ten choose five”:

C(10,5) = 10!/(10-5)!5! = 252. You could also think of it as 10x9x8x7x6/5! because you have 10 options for your first choice, times 9 options for your second choice, etc. and then you must divide by 5!, the number of ways that 5 items can be arranged if order matters, so as to count combinations, not permutations.

So out of 252 possible combos of 5 pairs of socks, how many include at least one pair of each color? Let’s list the 5 possible ways to get at least one pair of each color: 3White-1Black-1Red, 2White-2Black-1Red, 2White-1Black-2Red, 1White-3Black-1Red, 1White-2Black-2Red.

Now we need to count the number of ways we can get each of those outcomes. For our first outcome, there are multiple ways to choose 3 white pairs of socks from 5 possible white pairs, 1 black from 3 black pairs, and 1 red from 2 red pairs.

3White-1Black-1Red = C(5,3) x C(3,1) x C(2,1) = 10 x 3 x 2 = 60

2White-2Black-1Red = C(5,2) x C(3,2) x C(2,1) = 10 x 3 x 2 = 60

2White-1Black-2Red = C(5,2) x C(3,1) x C(2,2) = 10 x 3 x 1 = 30

1White-3Black-1Red = C(5,1) x C(3,3) x C(2,1) = 5 x 1 x 2 = 10

1White-2Black-2Red = C(5,1) x C(3,2) x C(2,2) = 5 x 3 x 1 = 15

Add those all up and we get 175 possible ways to get at least one pair of each color.

175/252 = 0.694 or 69.4%