What is the probability that you have picked two balls with different colours?
Initially I tried to solve this using sampling without replacement, although I wasn't entirely sure if replacement was a factor here as it wasn't explcitly stated. I didn't get very far, as I got confused with the number of combinations, how many balls were picked and how many would be left in each set etc.
As is typical for me when I get confused with a probability question, I wrote out the tuples from the sample space and simply counted – this gave me 72 individual tuples, 18 of which were two balls that were the same colour leaving 54 combinations that would yield two different coloured balls. Thus the probability is $\frac{3}{4}$
I then thought to apply the same logic using combinations and tried to figure out what the probability would be of picking two of the same, event S, coloured balls which gave me $$\mathbb{P}(S)=\frac{{3 \choose 2}{3\choose1}}{9 \choose 2} =\frac{1}{4}$$
Thus the event D which is picking two different coloured balls is $$\mathbb{P}(D)=1-\frac{{3 \choose 2}{3\choose1}}{9 \choose 2}=\frac{3}{4}$$
Is this the correct answer? If it is (or even if it isn't) can someone please explain how one would go about doing this using combinations without having to find the probability of the compliment of the event first.
Thanks!
Best Answer
The symmetry of the problem comes to your aid. You have the same chance of picking any color ball for the first one. Having picked that, there are two balls of the same color left and six balls of different colors, so the chance the second matches is $\frac 28=\frac 14$. If you started with differing numbers of balls of each color it would be more complicated and not too different from what you did. If we started with $b$ blue, $g$ green, and $r$ red balls, the chance you get two blue balls is $\frac {b(b-1)}{(b+g+r)(b+g+r-1)}$ with similar expressions for red and green. You can add them together to get the chance the two balls match.