Question

Suppose you have an urn (a bucket) with 10 green balls and 7 red balls. (a) suppose you draw 8 balls out of the bucket all at once. how many ways are there to draw 4 red balls and 4 green ones

Answer 1

Calculate the probability of drawing 4 red balls and then 4 green balls as shown; 

P(4R+4G)=($\frac{7}{17}$)($\frac{6}{16}$)($\frac{5}{15}$)($\frac{4}{14}$)($\frac{10}{14}$)($\frac{9}{13}$)($\frac{8}{12}$)($\frac{7}{11}$)

P(4R+4G)=0.00308 chance 

The denominators are decreasing as the balls get taken out. 

Hope I helped :)