Finding the probability that $2$ socks are the same color Question: A drawer contains $6$ blue socks and $4$ white socks. Two socks are chosen randomly without replacement. What is the probability that the $2$ socks are the same color?Should I approach this problem by adding the probabilities of selecting $2$ blue socks and selecting $2$ white socks? If so, is the formula unordered without replacement? Can somebody direct me with the right formula to solving this?
 A: We can approach this in a few ways. Think about ordering, or not; look for matched options or for mismatched options (and calculate). None are very difficult.
Following your proposed approach, the calculation would look like:
$$\frac6{10}\cdot \frac 59 + \frac 4{10}\cdot \frac 39$$
Following the ordered mismatch approach:
$$1-\left( \frac6{10} \cdot\frac 49 + \frac 4{10}\cdot \frac 69\right)$$
A non-ordered find-the-matches approach:
$$\frac{\binom 62+ \binom 42}{\binom {10}2} $$
A non-ordered find mismatches approach:
$$1-\frac{\binom 61 \binom 41}{\binom {10}2} $$
All give the same result.
A: 
Should I approach this problem by adding the probabilities of selecting $2$ blue socks and selecting $2$ white socks? 

Well, that should be "or" rather than "and", but that is the correct idea.

If so, is the formula unordered without replacement? Can somebody direct me with the right formula to solving this?

It is the probability for selecting $2$ from $6$ blue socks or $2$ from $4$ white socks, when selecting any $2$ from all $10$ socks.   These are disjoint events so the probability of the union is the sum of the probabilities (so, yes, you just add them).

 $$\dfrac{\dbinom{6}{2}+\dbinom 42}{\dbinom{10}2}~=~\dfrac{6\cdot 5+4\cdot 3}{10\cdot 9}$$

A suggested alternative is looking at it as the probability for not selecting one sock of each colour.

 $$1-\dfrac{\dbinom 61\dbinom 41}{\dbinom {10}2}~=~1-\dfrac{2\cdot 6\cdot 4}{10\cdot 9}$$ 

These are the same value. 
A: You are correct to add the probabilities of selecting $2$ blue socks and selecting $2$ white socks.
To find each of these probabilities, you would multiply the probability that sock $1$ is a given color times the probability of sock $2$ being that same color.  But since it is without replacement, remember that for the second pick there is one less sock of the color that was already picked and one less sock overall.
Therefore, the probability of getting two blue socks would be ${6\over 10}*{5\over 9}$ and the probability of getting two white socks would be ${4\over 10}*{3\over 9}$.  Add these together to get the total probability of getting two socks of the same color. 
A: Look at it this way: 
No matter what your first choice is, you can still get two socks of the same color. Therefore the probability of your first sock working is $1$. 
We note that the probability of you picking two socks the same color can be represented as such: 
$P(M)=P(C) \cdot P(M_{2})$, where $M$ is the event of drawing two matching socks (and therefore $P(M)$ is the probability of you getting two matching socks. $P(C)$ is likewise the probability of you drawing any sock (because any sock will work for the first one). $P(M_2)$ is the probability of another sock being drawn to match the first. 
$$P(M)=P(C) \cdot P(M_{2})=P(M_2)$$
$P(M_2)$ is easily calculatable. The probability of the second sock matching the first sock if the first sock is blue is $5/9$. The probability of the second sock matching the first sock if the first sock is white is $3/9$ or $1/3$. Therefore, we have: 
$$P(M)=P(M_2)=\frac13 \cdot P(W)+\frac59 \cdot P(B)$$
Where $P(W)$ is the probability of drawing a white as the first sock and $P(B)$ is the probability of drawing a blue as the first sock. 
Substitution gives: 
$$P(M)=P(M_2)=\frac13 \cdot P(W)+\frac59 \cdot P(B)=\frac13\cdot\frac25+\frac59\cdot\frac35=\frac{6}{45}+\frac{15}{45}=\frac{21}{45}=\frac7{15}$$
