I did complete the first part of the question however, I'm not exactly sure on how to do the second. Just correct the part of my formula would need to change. Thanks!
An urn has 13 balls that are identical except that 6 are white and 7 are red. A sample of 7 is selected randomly without replacement.
(a) What is the probability that exactly 5 are white and 2 are red?
$$\frac{C(6,5) \cdot C(7,2)}{C(13,7)}=\frac{126}{1716}$$
(b) What is the probability that at least 5 of the balls are white?
for this would I do something similar to: $$\frac{C(6,6) \cdot C(6,5)}{C(13,7)}$$
If you have any edits to make to the question to improve it or understand better, feel free to edit it
A) Okay.$$\mathsf P(W=5) ~=~ \dfrac{\binom 65 \binom 72}{\binom{13}{7}}$$ Similarly $$\mathsf P(W=6) ~=~ \dfrac{\binom 66 \binom 71}{\binom{13}{7}}$$
B) So, $\mathsf P(W\geq 5) = \mathsf P(W=5\cup W=6)$