Say you have a bag with 4 balls, 2 of which are red and 2 are blue. You select 2 balls randomly from the bag and you want to know the probability of selecting exactly 1 red ball.
Without replacement, there are two simple methods:
-
$\binom{2}{1}\frac{2}{4}\frac{2}{3} = \frac{2}{3}$
-
$\frac{ \binom{2}{1}\binom{2}{1}} {\binom{4}{2}} = \frac{2}{3}$
With replacement we can do this:
- $\binom{2}{1}\frac{2}{4}\frac{2}{4} = \frac{1}{2}$
My question: Is there a way I can find the probability of exactly 1 red ball with replacement using ONLY combinations in the numerator and denominators, like in example 2?
I tried using the $\binom{r+n-1}{r}$ formula:
$\frac{ \binom{2+1-1}{1}\binom{2+1-1}{1}} {\binom{4+2-1}{2}} = \frac{ \binom{2}{1}\binom{2}{1}} {\binom{5}{2}}$ but I got $\frac{2}{5}$ instead of the correct answer which is $\frac{1}{2}$. I'm curious to know.
Best Answer
With replacement we just have independent trials with success probability $\frac{2}{4} = \frac{1}{2}$. We do two trials, one of which must be a success (red) and one a failure. We have $\binom{2}{1}$ ways of choosing the trial with a success and all sequences with one sucess and one failure have probability $\frac12 \frac12$ by the product rule.
So generalising for $r$ reds out of $n$ trials in a bag with $k$ reds and $m-k$ blues (so $m$ balls total), we have $$P(r \text{ reds}) = \binom{n}{r}(\frac{k}{m})^r(\frac{m-k}{m})^{n-r}= \frac{\binom{n}{r}k^r(m-k)^{n-r}}{m^n}$$
while the corresponding probability without replacement equals, provided $m \le n, r \le k, n-r \le m-k$:
$$P(r \text{ reds}) = \frac{\binom{k}{r}\binom{m-k}{n-r}}{\binom{m}{n}}= $$