Unable to recreate the result

Question

I can't seem to figure out how to get the result shown below. Can someone please help me understand how to get this result?

$$P(red>0,black>0)=\sum_{r=1}^2\left(\sum_{b=1}^2\frac{\binom2r\binom2b\binom{26}{4-r-b}}{\binom{30}4}\right)=\frac{281}{5481}$$

(Original image here.)

So, step by step, here is my math:

1. The result of the first Sigma is $3$
2. The result of the second Sigma is also $3$
3. The numbers for the numerator are $2\cdot 2\cdot 325=\binom21\binom21\binom{26}2$, giving a result of $1300$
4. The result for the denominator is $27405=\binom{30}4$
5. Now, multiplying 3 times the fraction $= \frac{3\cdot1300}{1\cdot27405}=\frac{3900}{27405}$.
6. So, now all that is left is $3$ times the fraction $3900/27405$, which results in $\frac{3\cdot 3900}{1\cdot 27405} = \frac{11700}{27405}$.

It's obvious that somehow, somewhere my math computing the numerator is wrong. I can see that $5481$ is the result of dividing $27405$ by $5$, but how the numerator result is supposed to be $281$ completely eludes me.

Can someone please be so kind and tell me where I went wrong? Much appreciated!

Answer 1

The problem is that you’ve badly misunderstood the summation notation. $\sum_{r=1}^2$ is not $3$; in fact, it means nothing by itself. If $f$ is some function of $r$, then $\sum_{r=1}^2f(r)$ is meaningful: it means $f(1)+f(2)$. And in your case the function $f$ is given by

$$f(r)=\sum_{b=1}^2\frac{\binom2r\binom2b\binom{26}{4-r-b}}{\binom{30}4}\;,$$

so you want

$$\sum_{b=1}^2\frac{\binom21\binom2b\binom{26}{4-1-b}}{\binom{30}4}+\sum_{b=1}^2\frac{\binom22\binom2b\binom{26}{4-2-b}}{\binom{30}4}=\sum_{b=1}^2\frac{2\cdot\binom2b\binom{26}{3-b}}{\binom{30}4}+\sum_{b=1}^2\frac{1\cdot\binom2b\binom{26}{2-b}}{\binom{30}4}\;.$$

Simlarly, $\sum_{b=1}^2$ means nothing by itself, but

$$\begin{align*} \sum_{b=1}^2\frac{2\binom2b\binom{26}{3-b}}{\binom{30}4}&=\frac{2\binom21\binom{26}{3-1}}{\binom{30}4}+\frac{2\binom22\binom{26}{3-2}}{\binom{30}4}\\\\ &=\frac{2\cdot2\cdot\binom{26}2}{\binom{30}4}+\frac{2\cdot1\cdot\binom{26}1}{\binom{30}4}\;. \end{align*}$$

I’ll leave it to you to work out

$$\sum_{b=1}^2\frac{1\cdot\binom2b\binom{26}{2-b}}{\binom{30}4}$$

and finish the computation.