Probability of first and second drawn balls of the same color, without replacement I have an urn with 10 balls: 4 red and 6 white. What is the probability that the first two drawn balls have the same color?
Approach No. 1:
$|U| = 10! = 3628800$
White balls on first and second draw: on the first and second draw, I can pick any of the 6 white balls on the first draw, and 5 on the second draw. After that, I have 8 balls left, of which I don't care what order they are in. This will become $6\times5\times8\times7\times6\times5\times4\times3\times2\times1=1209600$.
$\left(\frac{1209600}{3628800} \right)=\frac{1}{3}$
If we have red balls on the first and second draw, then: I can pick any of the 4 red balls on the first draw, and on the second draw any of the left over 3. This will be$4\times3\times8\times7\times6\times5\times4\times3\times2\times1=483840$.
$\left(\frac{483840}{3628800} \right)=\frac{2}{15}$
The sum of these two will be: $\frac{1}{3}+\frac{2}{15}=\frac{7}{15}$
Approach number two:
Chance of a red ball on first draw: $\frac{4}{10}$. Chances of a red ball on second draw: $\frac{3}{9}.$
$\frac{4}{10}\times\frac{3}{9}=\frac{2}{15}$.
Chance of a white ball on first draw: $\frac{6}{10}$. Chances of a white ball on second draw: $\frac{5}{9}$.
$\frac{6}{10}\times\frac{5}{9}=\frac{5}{15}$.
$\frac{5}{15}+\frac{2}{15}=\frac{7}{15}$
What I want to know, is this answer correct? Somehow I can't get it straight in my head that this is the right answer, since the $8!$ seems a bit weird to me. I mean don't the chances of the other red balls in the case of red balls on first and second draw count?

S = {Set of all possible combination of 2 balls from the available 4 Red and 6 White Balls}
n(S) = $^{10} P_{2}$ ways = $10 \times 9 = 90$ (Since, 2 balls can be selected from $10$ in $^{10} P_{2}$ ways)
Now,
E = Both are of same colour
n(E) = $2$ Red balls can be selected from $4$ in $^4 P_{2}$ ways AND $2$ White balls can be selected from $6$ in $^6 P_{2}$ ways)    Total = $^4 P_{2} + ^6 P_{2} =(12 + 30) =42$ ways. 
P(E) = $\frac{n(E)}{n(S)} = \frac{42}{90} = \frac{7}{15}$.
