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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


  • 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?

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Do a tree, starting from "no socks", see in what fraction of the cases you get a white one, or a black one; from "one white" what are the options, and the same for "one black". – vonbrand Feb 3 '13 at 19:40
Your calculation assumes you put the first sock back in the drawer before selecting the second sock and is correct if that is the procedure. If you don't put the first sock back, after taking a black out you have four blacks out of eight, so the chance of two blacks is $\frac 59 \cdot \frac 48$ The calculation for white changes similarly. – Ross Millikan Jan 11 '14 at 19:31

$2$ 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 $\frac16+\frac5{18}=\frac49$


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 $\frac5{4+5}\frac{5-1}{4+5-1}=\frac5{18}$

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I dont understand how you get 6 ways and 36 ways in the beginning. Also, a minor thing, you addressed the socks as balls in the middle. – Ghost Feb 3 '13 at 15:27
@Ghost, please find the edited version. Btw, $\binom nr$ is the number of combinations for choosing $r$ elements from the $n$ elements. – lab bhattacharjee Feb 3 '13 at 15:33

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.

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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.

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Probability of first being block = 5/9 2nd being block = (5-1)/(9-1) Therefore both being block = 5/9 * 4/8 = 5/18

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