SOLUTION: I know this is a hypergeometric distribution. When I solve for P1 = success of daffodil and P2 = success for tulip...I get the same answer and now I am confused. Something seems o

Question 206723: I know this is a hypergeometric distribution. When I solve for P1 = success of daffodil and P2 = success for tulip...I get the same answer and now I am confused. Something seems off to me.
A homeowner plants 6 bulbs selected at random from a box containing 5 tulip bulbs and 4 daffodil bulbs. What is the probability that the homeowner planted 2 daffodil bulbs and 4 tulip bulbs?
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
ways to select 6 bulbs out of 9 ___ 9C6 = 84

ways to select 2 d-bulbs out of 4 ___ 4C2 = 6

ways to select 4 t-bulbs out of 5 ___ 5C4 = 5

ways to select 2 d-bulbs and 4 t-bulbs ___ 6 * 5 = 30

probability of 2 d and 4 t ___ (30 / 84) = (15 / 41) = .366 (approx)
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