So there are 5 red balls and 4 blue balls in an urn. We select two at random by putting them in a line and selecting the two leftmost. What's the probability both are different colors if the balls are numbered, and if the balls of the same color are indistinguishable? All arrangements have the same probability.
I don't think distinguishability of balls matters here, does it? All we care about is the color. So in both cases there's a 5/9 probability of red being leftmost, and a 4/9 probability of blue being leftmost. Given red is leftmost, there's a 4/8 chance that blue is next and given blue is first there's a 5/8 chance that red is next. So the probability for both numbered and unnumbered = (5/9)(4/8) + (4/9)(5/8). Is this correct, or am I missing something?
Best Answer
Distinguishability is not a property of the balls. It is the property of the observer. If there are balls of different colors arranged in boxes then certain arrangements will be distinguishable for those with a good eyesight and some arrangements will not be distinguishable for the colorblind.
In probability theory objects are called indistiguishable if the observer (experimenter) finds those arrangements equally likely that he can distinguish when, in theory, other observers with better eyes could distinguish further arrangements.
That is: "indistiguishability" is a misnomer.
So your answer is correct whether or not you are able to see the numbers on the balls. If you are then you would have more elementary events but you would have to unite some.
See again André Nocolas' answer.