BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

set of different flowers - probability

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 34
Joined: Mon May 07, 2012 10:59 am
Thanked: 2 times
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.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Data Sufficiency |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3835
Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members
Location: Milpitas, CA
Thanked: 1854 times
GMAT Score:770

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]
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 273
Joined: Tue Sep 21, 2010 5:37 amFollowed by: 74 members
Location: Durham, NC
Thanked: 154 times
GMAT Score:770

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
Whitney Garner
GMAT/GRE/EA Instructor & Anxiety/Accommodations Coach
www.whitneygarner.com

Contributor to Beat The GMAT!

Math is a lot like love - a simple idea that can easily get complicated :heart-eyes:
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members
Location: Vancouver, BC
Thanked: 5254 times
GMAT Score:770

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
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 118
Joined: Mon May 21, 2012 10:07 pmFollowed by: 4 members
Thanked: 23 times

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
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 34
Joined: Mon May 07, 2012 10:59 am
Thanked: 2 times

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
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1