harpritsn wrote:A florist has 2 azaleas, 3 buttercups, and 4 petunias. She puts two flowers together at random in a bouquet.
However, the customer calls and says that she does not want two of the same flower.
What is the probability that the florist does not have to change the bouquet?
Please help for this question.
source MGMAT
Here'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
