A bag contains 15 balls of the same shape and size. Of these, 9 balls are blue, and the remaining 6
balls are red. Suppose 7 balls are removed randomly (without replacement) from the bag, in such
a way that any 7 balls originally in the bag is equally likely to be the 7 balls that are removed from
the bag. What is the probability that the number of red balls removed from the bag is exactly 4?
I tried to figure out that the Sample space: drawing 2 balls. So we have a total of 15 balls and we can draw any 7 so the total possibilities are 15C7. Don't know how to proceed further
Best Answer
There are $15\choose 7$ ways to make the selections. Of these, there are ${9\choose 3}\cdot {6\choose 4}$ ways to choose exactly $4$ red balls. So...?