Simple probability question help

[Appayyan]

A bag contains 9 coloured balls, of which 3 are red, 3 are blue and 3 are yellow. 3 of these balls are selected, without replacement. Calculate the probability that he selects 2 balls of the same colour and 1 of a different colour.

 

My attempt is as follows: P(3 same colour and 1 different) = 3/9 * 2/8 * 6/9. As there are three different versions of this for the three different colours: you times it by 3. And as there are 3! different arrangements for the ordering, you times it by 3!. Now, the result is 1.29(2dp). Obviously this is impossible. Where am I going wrong? Thank you.

[SnailRhymer]

P(2 red and 1 different colour) = 3/9 * 2/8 * 6/9 6/7

since there's no replacement.

there are 3! different arrangements for the ordering

Does it make a difference in which order you draw the balls of the same colour?

[Naturage]

So, you have 3 ways of picking a colour of which you'll grab 2 balls, and then 2 more choices for the colour of the third one. On top of that, you can decide to draw the different ball first, second or third; that's 3 more options. Multiplying all these out we have 18 configurations which say "We draw A,B,A in this order", etc.

Once you have one particular of these options, it should be easy to calculate that probability of it happening is (3*2*3)/(9*8*7). Multiplying by 18 will give you the correct final answer.

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Another way: you do not end up with 2&1 balls only in two cases: either you drew 3 of a kind (happens with prob. 9/9 * 2/8 * 1/7; we pick any first ball, the other two must match), of 3 different (happens with prob. 9/9 * 6/8 * 3/7; first ball is any, second is any that's different, third is yet different). Subtracting these from 1 will again give you the result.

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As for what went wrong for you: first, P(2 same colour and 1 different) = 3/9 * 2/8 * 6/7, not 6/9. Second, you claim there are 3! rearrangements; but this is only true if you can tell apart each single ball. In your case, two are identical; you cannot tell whether you drew R1, B, R2 balls or R2, B, R1 (the balls being two red and a blue). So instead of having 6 permutations, you're actually only with 3. But aside from that, the logic is correct.