What's the probability if getting the same objects of the same colour? Q:A bag contains 5 black socks and 4 white socks. If 2 socks are picked randomly from it, find the probability of them being of the same colour.
What I've done is this:


*

*probability of first sock being black: 5/9 

*probability of first sock being white: 4/9 

*Probability of both being black is therefore, 5/9 * 5/9 = 25/81
similarly, 


*

*Probability of both being black is therefore, 4/9 * 4/9 = 16/81


And probability of any 2 socks being of the same color is then: 25/81  +   16/81

I dont think what I've done is right. How would this be done?
 A: $2$ white socks from $4$ white socks can be chosen in $\binom 42=\frac{4\cdot3}{2\cdot1}=6$ ways.
Any $2$ socks can chosen from $9$ socks in $\binom92=\frac{9\cdot8}{2\cdot1}=36$ ways.
So, the  probability of first two socks being white is  $\frac{\text{ the number of favourable cases }}{\text{  the number of possible cases }}=\frac{6}{36}=\frac16$
Similarly, the probability of first two socks being black is $\frac{\binom52}{\binom92}=\frac5{18}$
So, the probability of first two socks being of same colour is $\frac16+\frac5{18}=\frac49$

Alternatively,
The probability of first socks being white is $\frac4{4+5}=\frac49$
The probability of second socks being white with 1st one also white is $\frac{4-1}{4+5-1}=\frac38$
So, the probability of first two socks being white is $\frac49\cdot\frac38=\frac16$
Similarly, the probability of first two socks being black is $\frac5{4+5}\frac{5-1}{4+5-1}=\frac5{18}$
A: It's perhaps more intuitive to look at it in terms of permutations rather than combinations (imagine the socks being numbered so that they are uniquely distinguishable). 
There are 4 ways to choose a first white sock and 3 for the second (there are only 3 left after pulling one white one) = 12 ways to get 2 white socks.
Similalry there are 5 ways for a first black sock and 4 for a second = 20 ways to get 2 black socks.
In total, 32 ways to get socks the same colour. 
The total possibilities are 9 for the first sock and 8 for the second = 72. I.e. there are 72 different outcomes i.e. 'permutations' (with numbered socks, each of which is equally likely to be selected), so the probability of two the same colour = 32/72 = 4/9.
The proof using combinations embeds the fact that you don't care which is the first sock and which is the second, so that whith numbered socks White1 + White3 is considered the same as White3 + White1. So in calculating combinations the permutations are divided by 2, and since ALL permutations are divided by 2 you come to the same answer.
A: There are ${9\choose 2}=36$ ways to draw two socks from the bag. $5\cdot 4=20$ of these consist of one black and one white sock. It follows that there are $36-20=16$ good choices, resulting in a probability of ${16\over 36}={4\over 9}$ for such a choice.
A: Probability of first being block = 5/9
2nd being block = (5-1)/(9-1)
Therefore both being block = 5/9 * 4/8 = 5/18
