permutations n probabilty!MGMAT!

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?

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

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

Great solution, sana!

The great thing about counting questions is that you can typically use probability rules or counting methods.

Cheers,
Brent