theachiever wrote:A florist has 2 azaleas, 3 buttercups, and 4 petunias. She puts 2 flowers together at random in a bouquet. However, the customer calls and says that she does not want 2 of the same flower. What is the probability that the florist does not have to change the bouquet?
First, we can rewrite the question as
"What is the probability that the two flowers are different colors?"
Well, P(diff colors) = 1 -
P(same color)
Aside: let A = azalea, let B = buttercup, let P = petunia
P(same color) = P(both A's
OR both B's
OR both P's)
= P(both A's)
+ P(both B's)
+ P(both P's)
Now let's examine each probability:
P(both A's):
We need the 1st flower to be an azalea and the 2nd flower to be an azalea
So, P(both A's) = (2/9)(1/8) = 2/72
P(both B's)
= (3/9)(2/8) = 6/72
P(both P's)
= (4/9)(3/8) = 12/72
So,
P(same color) = (2/72) + (6/72) + (12/72) = 20/72 =
5/18
Now back to the beginning:
P(diff colors) = 1 -
P(same color)
= 1 -
5/18
= [spoiler]13/18[/spoiler]
Cheers,
Brent
