Drawing cards from 100 black and 100 white cards, what is the chance one is black and one is white? So, Given 200 cards, from which 100 are black and 100 are white, I'm trying to figure out what is the probability that I draw 2 cards, st exactly one is black and 1 is white.
so $P(A)=\frac{\binom{100}{1}\binom{100}{1}}{\binom{200}{2}}$
Is this correct?
If so, what is the probabilty that for TWO different draws, I get exactly 1 white and 1 black cards?
 A: Yes, correct.
The second answer is the following
$\frac{\binom{100}{1}\binom{100}{1}}{ \binom{200}{2}} \cdot \frac{\binom{99}{1}\binom{99}{1}}{ \binom{198}{2}} $
A: You did not mention how you get this result, but it may be interesting to note that there is an easy way to get the same result.
The first kid has a probability of one to first get a card of a given color, and then a probability of $\frac{100}{199}$ to get a card of the other color, as there remains $100$ cards of the good color, for a total of $199$ remaining cards.
If the first kid gets two cards of different colors, then $198$ cards remain, $99$ of each color.
Therefore, the second kid has a probability of $\frac{99}{197}$ to get two cards of the same color.
The global probability is then equal to
$$\frac{100}{199} \cdot \frac{99}{197}$$
A: I think you have to be more careful about what your question is.
The probability that exactly one child out of the 100 children will receive one black and one white is zero. The remaining cards can not be distributed so that all the remaining children have two of each colour. At least one other child must also receive one black and one white.
