Statistics-Math-Probability An urn is filled with 10 green balls, 4 red balls, and 2 orange balls.  Three balls are selected without replacement.  Calculate the following probabilities:

(a) P(at least one ball is red)
I think the answer is: (4/16)(3/15)(2/14) = .007 or 1/140

(b) P(exactly one ball is orange)
I think the answer is: (2/16)(14/15)(13/14) + (14/16)(2/15)(13/14) + (14/16)(13/15)(2/14) = 13/40 or .325

(c) P(at most one ball is green) 

(d) P(neither ball is red)
For part (a), I would calculate the probability that $no$ red ball is selected, and subtract the result from one.

For part (b), you are correct, but notice that your three terms are equal. Why is this? Try to find the pattern and generalize it.

For part (c), you can take the sum of $P$(exactly one ball is green), whose calculation mirrors that of part (b), plus $P$(no ball is green), whose calculation mirrors that of part (a). 

For part (d), you just need to find P(no ball is red), which you should be able to recycle from part (a).
Answer:

Total number of ways you can select 3 balls = ${16\choose3}$

Part a) $$=1-\frac{{12\choose3}}{{16\choose3}} = 0.6071$$

Part b)$$=\frac{{2\choose1}{14\choose2}}{{16\choose3}}=0.325$$

Part c) $$=\frac{{10\choose0}{6\choose3}+{10\choose1}{6\choose2}}{{16\choose3}}=0.3035$$

Part d) $$=\frac{{12\choose3}{4\choose0}}{{16\choose3}}=0.39285$$