Explanation We will solve this question in a moment. First let us understand one concept which is bit confusing.
Suppose there are 4 objects and I ask you to pick up 2 from them. In how many ways can you do that?
The answer is simple: ^{4}C_{2} = 6
Can I do like this:
Out of 4 given objects first I will pick one object(^{4}C_{1}). Then again I will pick one more object in the next draw from the remaining 3 objects(^{3}C_{1}).
This way I have selected two objects from fours objects and number of ways to do this is: ^{4}C_{1 }* ^{3}C_{1} = 12
Do you see in both the way I am selecting two objects from four objects but in first way of selection the result is 6 while in the other way it is 12. Why so ??
This is because in the second way when we are selecting two objects one by one, their order of selection is being counted in the result. Didn't understand ?
OK, let's understand this way: Suppose we mark all 4 objects from 1 to 4.
One way is: In the first draw we select object marked 1 and in the second draw we pick object marked 2.
The other way is: the first selected object is marked 2 and second object is marked 1.
We are selecting the same two objects but the number of way is 2 because their order of selection (1,2) and (2,1) made it happen so.
But our ultimate goal was to pick up two object and the order does not matter to us.
Similarly if you select 3 objects from these 4 objects you can have following orders for objects marked 1,2,3: (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1) = 6 ways= 3! ways. While it should be just 1 way and not 6 ways.
So we can conclude that we will have to divide the result by the factorial of the number of objects we are selecting.
Like in the first example when we are selecting 2 objects we will have to divide the result by 2! and in the second example when we are selecting 3 objects we will have to divide the result by 3!.
Now coming back to our given problem:
At least on pair of socks must be selected means either one pair or two pair.
So required probability will be: 1 - (no pair of socks are selected)
The way we can select no pair of socks is:
Select one sock from 20 sock in the first pick. In the second draw exclude the pair of the first selected sock and pick from the remaining 18 socks and so on.
= ^{20}C_{1 * }^{18}C_{1 * }^{16}C_{1 * }^{1}^{4}C_{1}/4! (4! because of the above explained concept)
So required probability: 1 - ((^{20}C_{1 * }^{18}C_{1 * }^{16}C_{1 * }^{1}^{4}C_{1}/4!)/^{20}C4) = 99/323