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I have an urn with b blue balls and r red balls, what is the probability that the two balls will have the same color. I need to solve this problem with replacement and without replacement. This is my attempt at a solution. Without replacement, we add the probabilities of the balls both being blue and the balls both being red. That is, the final probability would be $\frac{b(b-1)+r(r-1)}{br}$. With replacement, the final probability would be $\frac{b*b+r*r}{br}$. Is my logic correct in determining the probability?

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  • $\begingroup$ With replacement it would just be (b+r)(b+r)? $\endgroup$ – user288829 Jan 26 '16 at 17:29
  • $\begingroup$ Yes, for replacement it is $(b+r)(b+r)$. $\endgroup$ – André Nicolas Jan 26 '16 at 18:06
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No total ways $${b+r\choose 2}$$ so probability will be $\frac{(b)(b-1)+(r)(r-1)}{(b+r)(b+r-1)}$ same for other one

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