Ball Probability Question

Six balls are selected at random without replacement from an urn containing three white balls and five blue balls. Find the probability that two or three of the balls are white. (Round your answer to three decimal places.)

P(A): possible of no outcomes/total of no outcomes

8 Choose 5
3 Choose 2

P(2 white)=(3C2)(5C1)/8
P(3 white)=(3C3)(5C1)/8

Now for C do I use nPr or nCr? Also do I add or multiply them together to find the answer? Thanks.

The probability that 2 are white, then the other 4 must be blue:

\(\displaystyle \frac{\binom{3}{2}\binom{5}{4}}{\binom{8}{6}}\)

If 3 are white, then 3 must be blue:

\(\displaystyle \frac{\binom{3}{3}\binom{5}{3}}{\binom{8}{6}}\)

Add the two cases to find the probability that 2 OR 3 are white.