AIM TO CRACK GMAT wrote:A florist has 2 azaleas, 3 buttercups and 4 petunias. She puts two flowers together at a random in a bouquet. However, the customer calls and tells she does not want the two of d same flower. What is the probability that the florist doesnt have to change the bouquet?
[spoiler]13/18[/spoiler]
ere's an approach that does not use the complement.
To calculate P(2 diff colors), we need to consider 3 cases.
That is, P(2 diff colors) = P(azalea 1st then a different flower
OR buttercup 1st then a different flower
OR petunias 1st then a different flower)
= P(azalea 1st then a different flower)
+ P(buttercup 1st then a different flower)
+ P(petunia 1st then a different flower)
Let's examine each probability separately.
case 1: choose
azalea first, choose different flower second
The probability of choosing an azalea first is 2/9
Once we have selected an azalea first, there are 8 flowers remaining (1 A, 3 B's and 4 P's), and we must choose a different flower second.
So, the probability of choosing a different flower second is 7/8
So, P(azalea first and different flower second) = (2/9)(7/8) =
14/72
case 2: choose
buttercup first, choose different flower second
The probability of choosing an buttercup first is 3/9
Once we have selected an buttercup first, there are 8 flowers remaining (2 A's, 2 B's and 4 P's), and we must choose a different flower second.
So, the probability of choosing a different flower second is 6/8
So, P(buttercup first and different flower second) = (3/9)(6/8) =
18/72
case 3: choose
petunia first, choose different flower second
The probability of choosing an petunia first is 4/9
Once we have selected an petunia first, there are 8 flowers remaining (2 A's, 3 B's and 3 P's), and we must choose a different flower second.
So, the probability of choosing a different flower second is 5/8
So, P(petunia first and different flower second) = (4/9)(5/8) =
20/72
P(2 diff colors) =
(14/72) + (18/72) + (20/72)
= 52/72
= [spoiler]13/18[/spoiler]
Cheers,
Brent
