Probability question.Mismatch in answer in 2 diff approaches

[vmdb]

Question: A florist has 2 Azaleas, 3 Buttercups, and 4 Petunias. She put two flowers together at random in a bouquet. However a customer does not want two of the same flower. What is the probability that the florist does not have to change the bouquet ?

I tried answering the above problem in two ways.

1) Prob of having 1 azalea&1 buttercup in a bouquet OR 1 buttercup&1 petunia in a bouque OR 1 Petunia&1 azalea in a bouquet.

=> 2/9*3/8 + 3/9*4/8 + 4/9*2/8 = 26/72

2) 1 - (Prob of having 1 azalea&1azalea OR 1 buttercup&1 buttercup OR 1petunia&1 petunia)

=> 1 - (2/9*1/8 + 3/9*2/8 + 4/9*3/8) = 1 - 20/72 = 13/18


My question is why am I getting TWO different answers. They both have to be the same. 13/18 is the correct answer, as per the Guide. What is the mistake I 'm doing in my FIRST approach

[ace_gre]

Hi , Here is my take on the first approach..

2/9*3/8 *2+ 3/9*4/8 *2+ 4/9*2/8 *2

*2 because there are two ways of selecting one azalaa and one buttercup==>
1 azalea & 1 buttercup
1 buttercup & 1 azalea

Same for the other combinations. So 26/72 * 2 = 13/18.
Not an issue in second approach because both the flowers are of the same type...

Hope this helps!

[vmdb]

I didn't really understand, how a) picking first: Azalea & picking second: Buttercup differ from b)picking first: Buttercup & picking second: Azalea. No matter how one picks up, contents of the bouquet remains the same.

I was thinking abt it yesterday, got a different way of analyzing the problem

Prob [(Azalea first AND (Buttercup OR Petunia) second) OR (Buttercup first AND (Azalea OR Petunia) second) OR (Petunia first AND (Buttercup OR Azalea) second)]

That evaluates to

2/9*(3/8 + 4/8) + 3/9*(2/8 + 4/8) + 4/9*(2/8 + 3/8) = 13/18

Anything wrong in this approach.