probability – A bag contains 6 white balls, 5 black balls and 2 red balls.

A bag contains 6 white balls, 5 black balls and 2 red balls.
If two balls are drawn at random, what is the probability that neither of them are white?

For this question, the method that I used was to consider the four possible cases, BB, RR, BR, RB.

Therefore $$P = P(BB)+P(RR)+P(BR)+P(RB)=frac{5}{13} times frac{4}{12} + frac{2}{13} times frac{1}{12} + frac{5}{13} times frac{2}{12} + frac{2}{13} times frac{5}{12} = frac{7}{26} $$

which gives the correct answer, but I found there are two other ways to do this question, neither of which I understand.

Alternative method 1):

$P = P(text{red/black first pick}) times P(text{red/black second pick})
=frac{5+2}{13} times frac{(5+2)-1}{13-1} = frac{7}{26}$

This obviously kind of makes sense, but I’ve just never seen a question done this way before, I would really appreciate if someone could explain logic behind this in detail, and also point towards some textbooks that explain this part. (I already searched through the textbook I have, but I haven’t seen them use this method in a worked example..)

Alternative method 2)
$$P = frac{{}_5C{}_2 + {}_2C{}_2 + {}_5C{}_1 times {}_2C{}_1}{_{13}C{}_2}=frac{7}{26}$$

What I don’t understand about this method is why we didn’t view BR and RB as two different cases, i.e., why we didn’t do

$$P = frac{{}_5C{}_2 + {}_2C{}_2 + {}_5C{}_1 times {}_2C{}_1 + {}_2C{}_1 times {}_5C{}_1 }{_{13}C{}_2} $$

Thank you so much for the explanation..