Monco wrote:
A garden center has seven female and eight male holly plants, but they aren’t labeled as such. If a customer randomly purchases three plants, what is the probability of getting both male and female plants (thus holly berries, eventually)? Write your answer as a reduced fraction.
I was wondering why my approach doesn't work to get the right answer.
Let's say he picks 2x male and 1x female, this gives (8/15)*(7/14)*(7/13) or he picks 2x female and 1x male which gives (7/15)*(6/14)*(8/13), but this doesn't result in the right answer. Why not? Isn't this the same as saying 1 - probability all male + probability all female?
The right answer should be 4/5.
Let's examine your first set of results.
Let's say he picks 2x male and 1x female, this gives (8/15)*(7/14)*(7/13)Your calculations cover only 1 possible way to get 2 males and 1 female.
That is P(1st plant is male AND 2nd plant is male AND 3rd plant is female) = (8/15)(7/14)(7/13)
You haven't included 2 other cases that yield 2 males and 1 female:
case 2) 1st plant is male AND 2nd plant is female AND 3rd plant is male
case 3) 1st plant is female AND 2nd plant is male AND 3rd plant is male
Fortunately, the two additional cases each have a probability of (8/15)(7/14)(7/13)
So, P(2 males and 1 female) = (8/15)(7/14)(7/13) + (8/15)(7/14)(7/13) + (8/15)(7/14)(7/13)
= 3 x (8/15)(7/14)(7/13)
The same applies to the second part of your solution (2 females and 1 male)
Cheers,
Brent