<s>[INST] A box has $4$ red and $20$ white balls. A person takes $10$ balls. What is the probability that all or none of the red balls were taken? A box has $24$ balls, $4$ red and $20$ white. One person takes $10$ balls and the second the remaining $14$. What is the probability that one of the two people picked up the $4$ red ones?
I don't understand why is this correct.
$$\frac{{20 \choose 6}+{20 \choose 10}}{24 \choose 10} $$
 [/INST] You can use the Hypergeometric distribution. 
a) The probability that the first person pick up 4 red ones is $\frac{\binom{4}{4}\cdot \binom{20}{6}}{\binom{24}{10}}$
Then the (conditional) probability that the second person does not pick up 4 red ones is just $1.$ 
b) The probability that the first person does not pick up 4 red ones is $\frac{\binom{4}{0}\cdot \binom{20}{10}}{\binom{24}{10}}$
The (conditional) probability that the second person pick up 4 red one is $\frac{\binom{4}{4}\cdot \binom{10}{10}}{\binom{14}{14}}=1$
So in total we have 
$$\frac{\binom{4}{4}\cdot \binom{20}{6}}{\binom{24}{10}}+\frac{\binom{4}{0}\cdot \binom{20}{10}}{\binom{24}{10}}=\frac{ \binom{20}{6}}{\binom{24}{10}}+\frac{ \binom{20}{10}}{\binom{24}{10}}=\frac{\binom{20}{6}+ \binom{20}{10}}{\binom{24}{10}}$$
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