# Probability of picking balls with same color with replacement and without replacement

This is one of our probability exercise:

We have an urn with m green balls and n yellow balls. Two balls are drawn at random. What is the probability that the two balls have the same color?

(a) Assume that the balls are sampled without replacement.

(b) Assume that the balls are sampled with replacement.

(c) When is the answer to part (a) larger than the answer to part (b)? Justify your answer. Can you give an intuitive explanation for what the calculation tells you?

For (a), I think there are only two possible outcomes of this experiment, which are either two green balls or two yellow balls. Since they are independent on each other, so we can use the addition rule. Since it is without replacement, so we can directly use combinations. $$P(\text{same color}) = \frac{\binom{m}{1}\times \binom{m-1}{1} + \binom{n}{1} \times \binom{n-1}{1}}{\binom{m + n}{2}}$$

The difference between replacement and no replacement is how the candidate pool changes after each action. With replacement, we don't need to modify the number of candidates anymore. $$P(\text{same color}) = \frac{\binom{m}{1}\times \binom{m}{1} + \binom{n}{1} \times \binom{n}{1}}{(m+n)^2}$$

However, I have no clues how to solve (c). I tried to use algebra to simplify my answer for (a) and (b), but it doesn't help.

Any hints or suggestions would be appreciated!

• Double check your formula for part (a). e.g. set $m=n=2$. What should the answer be? – John M Jun 15 '16 at 13:00
• @johnM Thank you, I see now. It should be $\frac{\binom{m}{2}+ \binom{n}{2} }{\binom{m + n}{2}}$ instead. I am a little bit confused, say we pick one ball from m, then second ball from (m-1). Why they are different by thinking choosing twice (1 ball and another ball) and choosing once (2 balls)? – Jay Wong Jun 15 '16 at 13:37
• If you make an unordered selection in the denominator, you must also make an unordered selection in the numerator. That yields the probability $$\frac{\binom{m}{2} + \binom{n}{2}}{\binom{m + n}{2}}$$ If you wish to make an ordered selection in the numerator, you must also make an ordered selection in the denominator. That yields the probability $$\frac{m(m - 1) + n(n - 1)}{(m + n)(m + n - 1)}$$ As you can check, the results are equal. What you cannot do is make an ordered selection in the numerator and an unordered selection in the denominator. – N. F. Taussig Jun 15 '16 at 13:45
• @N.F.Taussig Thank you! It really helps me a lot. – Jay Wong Jun 15 '16 at 14:29

In answer to part c), if you write, in the form of an inequality in $m$ and $n$, "answer to part(a) > answer to part(b)", this inequality reduces to the statement $$mn<0$$ which is clearly never true.