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set of different flowers - probability

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set of different flowers - probability

by kunalkulkarni » Mon Oct 08, 2012 8:01 pm
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?
- from MGMAT
Please help. having a hard time to understand the solution.
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by Anurag@Gurome » Mon Oct 08, 2012 8:17 pm
kunalkulkarni 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?
- from MGMAT
Please help. having a hard time to understand the solution.
Probability that the florist does not have to change the bouquet = 1 - (probability that the two flowers are azaleas + probability that the two flowers are buttercups + probability that the two flowers are petunias)
= 1 - [(2/9 * 1/8) + (3/9 * 2/8) + (4/9 * 3/8)]
= 1 - [20/72]
= 1 - 5/18
= [spoiler]13/18[/spoiler]
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by Whitney Garner » Mon Oct 08, 2012 8:30 pm
kunalkulkarni 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?
- from MGMAT
Please help. having a hard time to understand the solution.
Hi kunalkulkarni!

You can also solve this problem using combinatorics (if you feel more comfortable with those). The probability of NOT having to change = 1-Pr(Having to Change) = 1-Pr(2 same flower)

To get the probability of 2 of the same flower, we need the number of combinations that result in the same flowers divided by the total number of possible pairs.

The total pairs is easier - it is just 9 choose 2 or 9!/(7!2!) = (9*8)/2 = 36.
The number of pairs of 2 azaleas from 2 is just 2c2 = 2!/2! = 1
The number of pairs of 2 buttercups from 3 is 3c2 = 3!/(2!1!) = 3
The number of pairs of 2 petunas from 4 is 4c2 = 4!/(2!2!) = (4*3*2)/4 = 6

So there are 10 pairs we don't want out of 36 total - that is a probability of 10/36 = 5/18, so the probability of NOT getting those is 1-5/18 = 13/18.

Hope this helps!
:)
Whit
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by Brent@GMATPrepNow » Mon Oct 08, 2012 9:18 pm
kunalkulkarni 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?
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
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by das.ashmita » Mon Oct 08, 2012 9:20 pm
Hi Kunal

A li'l longer approach, but it never fails me :)

azaleas(A), buttercups(B), and petunias(P)

Total no. of ways to create a bouquet of 2 flowers = 9C2 = 36
No. ways to create a bouquet with 2 different type of flowers = AB + AP + BP
= 2C1.3C1 + 2C1.4C1 + 3C1.4C1 = 26

Probability that the florist does not have to change the bouquet = (AB + AP + BP)/Total = 26/36 = 13/18
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by kunalkulkarni » Tue Oct 09, 2012 11:27 am
Wow, I never imagined it could be this easy!
I guess the reason I didn't understand it earlier was perhaps I have a hard time with "combinatorics"
the other approaches are very simple and they make the problem look dead simple.
Thank you 'GMAT-Gurus'!

Kunal
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