Probability (through combinations with replacement method) An urn contains 12 chips: 4 blue chips numbered 1 through 4, 4 green chips numbered 1 through 4, 4 red chips numbered 1 through 4. If two chips are to be drawn at random with replacement, what is the probability that both chips are exactly the same?
I understand that the answer is 12 x (1/12 x 1/12) = 1/12
but I would like know if it is possible to use the nCr - combinations method (with replacement) to derive the same answer
An example would be this
I would be very grateful to receive any constructive feedback regarding my question. Thank you!
 A: If by "combinations with replacement" you mean stars-and-bars, the number of ways of distributing identical balls into distinct bins, no.  Some of these outcomes you would be counting are not equally likely to occur.  It is far more likely to get one green chip and one blue chip in any order than it is to get two green chips.  Remember that to use $Pr(A)=\dfrac{|A|}{|S|}$ where $S$ is the sample space, that requires that the sample space be equiprobable meaning that each outcome in the sample space is equally likely to occur.  There are two outcomes when playing the lottery, winning and losing, but you don't win the lottery with probability $\frac{1}{2}$.
Now, you could rephrase the solution as a single fraction and use counting techniques, letting the sample space be the number of ways to select two chips in sequence.  There are $12\times 12$ ways in which you can select two chips in sequence and as per our usual assumptions about selecting objects with replacement these ways are equally likely to occur.  The numerator then being $12\times 1$, counting the number of ways we can select a chip and then select the same chip again.  This gives us a probability of $\dfrac{12\times 1}{12\times 12} = \frac{1}{12}$, same as before.
