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.
$$\frac{{20 \choose 6}+{20 \choose 10}}{24 \choose 10} $$
 A: Because $\tfrac{\binom 44\binom{20}6\binom{14}{14}}{\binom{24}{10}\binom{14}{14}}$ is the probability that the first person select 4 red and 6 white and the second person receives the remaining 14 white, while $\tfrac{\binom{20}{10}\binom{4}{4}\binom{10}{10}}{\binom{24}{10}\binom{14}{14}}$ is the probability that the first person selects 10 white and second person select 4 red and 10 white.
$$~\\\dfrac{\binom 44\binom{20}6\binom{14}{14}+\binom{20}{10}\binom{4}{4}\binom{10}{10}}{\binom{24}{10}\binom{14}{14}}=\dfrac{\binom{20}6+\binom{20}{10}}{\binom{24}{10}}$$
A: 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 
$$\frac{\binom{4}{4}\cdot \binom{20}{6}}{\binom{24}{10}}+\frac{\binom{4}{0}\cdot \binom{20}{10}}{\binom{24}{10}}=\frac{ \binom{20}{6}}{\binom{24}{10}}+\frac{ \binom{20}{10}}{\binom{24}{10}}=\frac{\binom{20}{6}+ \binom{20}{10}}{\binom{24}{10}}$$
A: Firstly, note that the events are mutually exclusive events. Therefore, the probability overall is the sum of the probability of the first person and second person selecting 4 red balls.
The number of ways the first person selects the 4 red balls is the same as the number of ways they can select the remaining six as white balls (they have 10 chances in total) from the 20 white balls available. This can be done in $24-4\choose{6}$ ways
The number of ways the second person selects the 4 red balls is the same as the number of ways they can select the remaining ten as white balls (they have 14 chances on total) from the 20 white balls available. This can be done in $24-4\choose{10}$ ways.
The total number of ways to split the 24 balls is $24\choose{10}$
Therefore, the probability  is what you put down.
A: There are two ways for a "good event" to occur in this probability space: either player one picks up all 4 red balls (first type of good event), or player 1 picks up none of the red balls (second type of good event; equivalent to saying player 2 picks up all red balls). 
The total number of ways that the first type of good event can occur is the number of ways we can pick 10 - 4 = 6 balls amond 24 - 4 = 20 balls after we fix that player one has selected the 4 red balls. In other words, there are $20 \choose 6$ ways the first type of good event can occur.
The total number of ways that the second type of good event may occur is the number of ways player one can pick all of his balls exclusively from the white balls. Since there are 20 white balls and player 1 picks 10 balls in total, this gives us $20 \choose 10$ ways for the second type of good event to occur.
Since the two "good" event spaces are mutually exclusive (player 1 can't pick all the red balls and none of the red balls at the same time), we can add them up obtain the total number of good event (the numerator in the solution you provided), and divide this by the size of the probability space (the total number of ways player 1 can pick 10 balls from 24, which is your denominator term).
Does this answer your question?
