# Question: Odd or Even

## Comment on Odd or Even

### Say there are 5 matching

Say there are 5 matching pairs of socks and a person picks two random NON-MATCHING SINGLE socks out of a drawer (the draw is consecutive without replacement) ..what would the probability of that happening? Was thinking that you can pick any sock on the first draw which is 10/10 and then on the second draw you just can't get the same color sock so it would be 8/9. so 10/10 X 8/9 = .88 %
though not sure

### Perfect!

Perfect!

P(no matching pair) = P(select ANY sock as 1st selection AND select a NON-MATCHING sock as 2nd selection)

= P(select ANY sock as 1st selection) x P(select a NON-MATCHING sock as 2nd selection)

= 1 x 8/9

= 8/9

ASIDE: For others following along, the second probability is 8/9, because once we've selected the 1st sock, there are 9 socks remaining. Of these 9 socks, 1 sock matches the 1st sock selected, and the other 8 socks are NOT a match.

### Hi Brent,

Hi Brent,
If it were a counting question asking how many ways we can select two non matching sock, how would we go about it using FCP ?

### Here's one approach:

Here's one approach:

First off, let's say that, for each pair of matching socks, we have a left sock and a right sock.

STAGE 1: Choose 2 colors
There are 5 colors. Since the order in which we select the socks does not matter, we can use combinations.
We can select 2 colors from 5 colors in 5C2 ways (10 ways)

ASIDE: Now, from each color, we'll select either a left sock or a right sock

STAGE 2: From one color, choose a left or right sock
This can be accomplished in 2 ways

STAGE 3: From the other color, choose a left or right sock
This can be accomplished in 2 ways

TOTAL number of ways to select 2 NON-MATCHING socks = (10)(2)(2)
= 40

For "fun", let's also determine the denominator.

TOTAL number of ways to select ANY 2 socks = 10C2 = 45

So, P(no matching pair) = 40/45 = 8/9

Cheers,
Brent