What's the probability that I draw at least 1 white card when drawing 3 cards from 3 decks of 15 cards, 2 of which are white? I haven't seen quite this scenario on a card drawing problem on here. I'm trying to figure out the probabilities for a card game I'm developing. There are 3 separate decks with 15 cards each. In each deck there are 2 'white cards' let's say and we're interested in drawing those. So if I draw 3 cards from each of the decks, then what is the probability that I draw at least 1 'white card' and not one of the other 13?
I think I calculated the probability for one of the decks so I'll put my work here for someone to check it.
Probability of drawing at least 1 white card when drawing 3 cards from one deck:
First, I computed the probability of drawing exactly 1 white card when drawing 3 cards from one deck, which is as follows.
$P(W_1) =  (_3C_1) \left(\frac{2}{15}\right)^1 \left(\frac{13}{15}\right)^2$
$P(W_1) = 0.3004$
Then, I computed the probability of drawing 2 white cards when drawing 3 cards from the deck.
$P(W_2) = (_3C_2)\left(\frac{2}{15}\right)^2 \left(\frac{13}{15}\right)^1$
$P(W_2) = 0.0462 $
So then the probability of drawing at least 1 white card is,
$P(W)= P(W_1) + P(W_2) = 0.3466$
So how do I go about incorporating the other 2 decks into my equation? What's the probability of drawing a white card when drawing 3 cards from each deck? Thanks.
 A: Notice that the probability of drawing at least one white card is the same as $1$ minus the probability of drawing no white cards:
$$
P\left(\text{at least $1$ white card}\right) = 1 - P\left(\text{no white cards}\right)$$

The probability that no white cards are drawn when taking three cards from a single deck is:
$$P\left(\text{no white cards from a single deck over three draws}\right) = \left(\frac{13}{15}\right)\left(\frac{12}{14}\right)\left(\frac{11}{13}\right)$$

In order to get no white cards over all nine draws, it's necessary to get no white cards on the three draws from each of the three decks:
$$P\left(\text{no white cards}\right) = \left(\left(\frac{13}{15}\right)\left(\frac{12}{14}\right)\left(\frac{11}{13}\right)\right)^3$$

Therefore,
\begin{align*}
P\left(\text{at least $1$ white card}\right) &= 1 - P\left(\text{no white cards}\right)\\\\
&= 1 - \left(\left(\frac{13}{15}\right)\left(\frac{12}{14}\right)\left(\frac{11}{13}\right)\right)^3 = \boxed{\frac{32227}{42875}} \approx 0.75165
\end{align*}
A: So I was messing around with it more and talking to a friend about it. Can anyone tell me if this is correct...
$P(W) = 1 - P(W')$ 
$ = 1 - \left(\frac{^{13}C_3}{^{15}C_3}\right)^3 $
Which comes out to $P(W) = 0.752$
It just seems quite high to me.
