# A bag contains 15 balls of the same shape and size. Of these, 9 balls are blue...

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

• The next thing to consider is the number of ways the 7 balls can consist of 4 red and 3 blue. Nov 3, 2018 at 18:28
• how should i do that Nov 3, 2018 at 18:32

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...?