6. A bag contains 5 red balls, 6 black balls, and 7 white balls. Now, 15 balls are drawn from the bag. The probability that exactly 3 of the drawn balls are red is ( ).
(A) $\frac{6}{65}$
(B) $\frac{65}{408}$
(C) $\frac{13}{816}$
(D) $\frac{13}{4896}$
6.B.

There are a total of $5+6+7=18$ balls in the bag. Drawing 15 balls is equivalent to leaving 3 balls in the bag. Considering that the balls are indistinguishable in order, the number of different drawing results is $\frac{18 \times 17 \times 16}{3 \times 2 \times 1}=816$.

If exactly 3 red balls are drawn, we first need to draw 3 red balls from the 5 red balls, which has $\frac{5 \times 4 \times 3}{3 \times 2 \times 1}=10$ possible outcomes; then, we need to draw $15-3=12$ balls from the remaining 13 balls (6 black balls and 7 white balls), which means leaving 1 black or white ball in the bag, with 13 different outcomes. Therefore, the number of drawing results where exactly 3 red balls are drawn is $10 \times 13=130$.
Thus, the probability of drawing exactly 3 red balls is
$$
\frac{130}{816}=\frac{65}{408} .
$$