# Math Help - Probabilities...

1. ## Probabilities...

I thought this was gonna be easy...but I'm stumbling a little bit:

While dressing in the dark, Joe selects 8 socks at random from a drawer containing 10 different pairs. What is the probability that at least one pair of socks match?
I don't want anyone to solve this, but rather, I'd like to see if I'm doing this right.

Let $X$ represent the number of matching pairs. We are looking for $P(X\geq1)$. I figured that $P(X\geq1)=1-P(X=0)$, where $P(X=0)$ is the probability of none matching.

Since there are 20 different socks, and 2 are needed to make a pair (also noting that the pairs are different), I think that $P(X=0)=\frac{\displaystyle\binom{2}{1}^8\binom{2}{ 0}^2}{\displaystyle\binom{20}{8}}$

Thus, I get $P(X\geq1)=1-\frac{\displaystyle\binom{2}{1}^8\binom{2}{0}^2}{\ displaystyle\binom{20}{8}}\approx\boxed{.998}$

Does this look right?

I'd appreciate any help!!

2. Originally Posted by Chris L T521
Since there are 20 different socks, and 2 are needed to make a pair (also noting that the pairs are different), I think that $P(X=0)=\frac{\displaystyle\binom{2}{1}^8\binom{2}{ 0}^2}{\displaystyle\binom{20}{8}}$

Thus, I get $P(X\geq1)=1-\frac{\displaystyle\binom{2}{1}^8\binom{2}{0}^2}{\ displaystyle\binom{20}{8}}\approx\boxed{.998}$

Does this look right?
Not quite: on the numerator, you count the number of ways to choose socks in a given sequence of 8 pairs. As a consequence, you should multiply it by the number of ways to choose 8 pairs among 10 (like, first we choose the (unordered) pairs, and then the left/right sock in each pair).