probability

6 balls are selected at random without replacement from an urn containing 3 white balls and 5 blue balls. Find the probability that 2 or 3 of the balls are white.

C(13,4) divided by C(52,4)
C(n,r)=n!/(n-r)!

13!/(4!9!)
divided by
52!/(4!48!)

Why isnt this right?

Where did you get 52 and 13 from?. There are only 8 balls total.

It would appear you are mixing this up with a card problem.

The reason it isn't right is because you are not in the same Universe as the balls and the urn.


Find the probability for 2 whites and 3 whites, then add them.

\(\displaystyle \text{2 white}=\frac{\binom{3}{2}\cdot \binom{5}{4}}{\binom{8}{6}}\)

\(\displaystyle \text{3 white}=\frac{\binom{3}{3}\cdot \binom{5}{3}}{\binom{8}{6}}\)

add the two cases for the probability that 2 OR 3 are white.

So for 2 white: I got 720/20160 and for 3 white: I got 360/20160
so when I add 720 and 360=1080. 1080/20160 but this isn't right....why not?