Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 30 May 2017, 02:47

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

A florist has 2 azaleas, 3 buttercups, and 4 petunias. She

Author Message
TAGS:

Hide Tags

Senior Manager
Joined: 28 Aug 2010
Posts: 255
Followers: 6

Kudos [?]: 656 [0], given: 11

A florist has 2 azaleas, 3 buttercups, and 4 petunias. She [#permalink]

Show Tags

22 Jan 2011, 09:25
4
This post was
BOOKMARKED
00:00

Difficulty:

55% (hard)

Question Stats:

68% (02:38) correct 32% (02:13) wrong based on 182 sessions

HideShow timer Statistics

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?

A. 5/18
B. 13/18
C. 1/9
D. 1/6
E. 2/9
[Reveal] Spoiler: OA

_________________

Gmat: everything-you-need-to-prepare-for-the-gmat-revised-77983.html
-------------------------------------------------------------------------------------------------
Ajit

Last edited by Bunuel on 17 Nov 2012, 05:36, edited 1 time in total.
Renamed the topic and edited the question.
Math Expert
Joined: 02 Sep 2009
Posts: 39069
Followers: 7760

Kudos [?]: 106607 [1] , given: 11630

Show Tags

22 Jan 2011, 09:34
1
KUDOS
Expert's post
4
This post was
BOOKMARKED
ajit257 wrote:
A florist has 2 azaleas, 3 buttercups, and 4 petunias. She puts 2 flowers together at random in a bouquet. However customer calls and says she does not want two of the same flower. What is the probability that the florist does not have to change the bouquet ?

Let's count the probability of the opposite event and subtract it from 1. Opposite event would be that the florist made a bouquet with two of the same flower: $$\frac{C^2_2+C^2_3+C^2_4}{C^2_{9}}=\frac{10}{36}$$ --> $$P=1-\frac{10}{36}=\frac{26}{36}=\frac{13}{18}$$

_________________
Senior Manager
Affiliations: SPG
Joined: 15 Nov 2006
Posts: 327
Followers: 16

Kudos [?]: 752 [1] , given: 28

Show Tags

05 Feb 2011, 10:21
1
KUDOS
1
This post was
BOOKMARKED

total possibilities of selecting 2 flowers from 9 = 9*8= 72
position doesn't matter i.e. {AB} & {BA} mean the same. so total cases = $$\frac{72}{2} = 36$$

total cases where both flowers are same.

{AA} = 2*1= 2
{BB} = 3*2= 6
{PP} = 4*3= 12
= 2+6+12= 20

position doesn't matter. so total cases $$= \frac{20}{2} = 10$$

Ans $$= 1 - \frac{10}{36} = \frac{13}{18}$$

_________________

press kudos, if you like the explanation, appreciate the effort or encourage people to respond.

Intern
Joined: 24 Apr 2013
Posts: 16
Location: Russian Federation
Followers: 0

Kudos [?]: 20 [1] , given: 13

Pls help with explanation for this problem from MGMAT Strategy Guide 5 [#permalink]

Show Tags

02 Sep 2014, 00:00
1
KUDOS
What we have: 2A, 3B, 4P
Total: 9 flowers

Prob of selecting 2A: (2/9)(1/8)=1/36
Prob of selecting 2B: (3/9)(2/8)=1/12=3/36
Prob of selecting 2P: (4/9)(3/8)=1/6=6/36

1/36+3/36+6/36=10/36=5/18

1-(5/18)=13/18
Math Expert
Joined: 02 Sep 2009
Posts: 39069
Followers: 7760

Kudos [?]: 106607 [1] , given: 11630

Re: A florist has 2 azaleas, 3 buttercups, and 4 petunias. She [#permalink]

Show Tags

02 Sep 2014, 03:09
1
KUDOS
Expert's post
arpshriv 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?

Merging similar tropics. please refer to the discussion above.

_________________
Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 918
Followers: 36

Kudos [?]: 516 [1] , given: 5

Re: A florist has 2 azaleas, 3 buttercups, and 4 petunias. She [#permalink]

Show Tags

18 Apr 2017, 16:44
1
KUDOS
Expert's post
ajit257 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?

A. 5/18
B. 13/18
C. 1/9
D. 1/6
E. 2/9

We can use the following formula:

1 = P(getting two of the same flower) + P(not getting two of the same flower)

Let’s determine the probability of getting two of the same flower.

P(2 azaleas) = 2/9 x 1/8 = 2/72

P(2 buttercups) = 3/9 x 2/8 = 6/72

P(4 petunias) = 4/9 x 3/8 = 12/72

The probability of getting two of the same flower is, therefore, 2/72 + 6/72 + 12/72 = 20/72 = 5/18

Thus, the probability of not getting two of the same flower is 1 - 5/18 = 13/18.

_________________

Jeffery Miller

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2013
Followers: 163

Kudos [?]: 1832 [0], given: 376

Show Tags

05 Feb 2011, 10:37
##2 azaleas(A), 3 buttercups(B), and 4 petunias(P)##

To select two flowers, each from a different type is:
1A and 1P
or
1B and 1P
or
1A and 1B

We can select 1A from 2 in $$C^2_1$$ ways
We can select 1B from 3 in $$C^3_1$$ ways
We can select 1P from 4 in $$C^4_1$$ ways

P(2 flowers, each of different type) is

$$\frac{C^2_1*C^3_1+C^2_1*C^4_1+C^3_1*C^4_1}{C^9_2}$$

$$\frac{2*3+2*4+3*4}{9*4}$$

$$\frac{6+8+12}{36}$$

$$\frac{26}{36}$$

Ans: $$\frac{13}{18}$$
_________________
Manager
Joined: 12 Oct 2012
Posts: 129
WE: General Management (Other)
Followers: 1

Kudos [?]: 54 [0], given: 198

Florist 2azaleas, 3 buttercups and 4 petunias [#permalink]

Show Tags

16 Nov 2012, 20:19
dimri10 wrote:
2 azaleas, 3 buttercups, and 4 petunias for total of 9:
same flower:
2 azaleas- 2/9*1/8 of choosing the same flower.
3 buttercups- 3/9*2/8
4 petunias - 4/9*3/8
2/72+6/72+12/72=20/72 Probability to chhose the same flower.

we want the probability of not choosing so 1-20/72=52/72=26/36=13/18

Would someone please explain why do we multiply by 1/8, 2/8, 3/8?
Intern
Joined: 02 Mar 2015
Posts: 10
Followers: 0

Kudos [?]: 9 [0], given: 0

A florist has 2 azaleas, 3 buttercups, and 4 petunias. She puts... [#permalink]

Show Tags

02 Sep 2015, 09:49
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?

1. Counting manually (by drawing a table) all possible combinations of bouquets with two same flowes
2. Counting all possible combinations of two flower bouquets by using anagram grid
3. Making probability fraction of amount of possible bouquets with 2 same flowers to all possible bouquets (two same flower and two different flower bouquets)
4. Subtracting upper probability fraction (something we have to exclude) from 1.

[Reveal] Spoiler:
I don't understand the solution method being used here. I translated the question being asked as "what is the probability that the first two flower bouquet that the florist picked was NOT a bouquet of two same flowers, meaning what is the probability that the first two flower bouquet that the florist picked was either AB OR BP OR AP?". Basically I used probability tree to solve this but I got different answer : 13/36.

AB = One flower is Azalea AND second flower is Buttercup
2/9 x 3/8 = 1/12

OR

BP = one flower is buttercup and second is petunia
3/9 x 4/8 = 1/6

OR

AP = one flower is azalea and second is petunia
2/9 x 4/8 = 1/9

Final answer = 1/12 + 1/6 + 1/9 = 13/36

How is my thinking wrong?
Math Expert
Joined: 02 Sep 2009
Posts: 39069
Followers: 7760

Kudos [?]: 106607 [0], given: 11630

Re: A florist has 2 azaleas, 3 buttercups, and 4 petunias. She [#permalink]

Show Tags

02 Sep 2015, 09:56
MariaVorop 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?

1. Counting manually (by drawing a table) all possible combinations of bouquets with two same flowes
2. Counting all possible combinations of two flower bouquets by using anagram grid
3. Making probability fraction of amount of possible bouquets with 2 same flowers to all possible bouquets (two same flower and two different flower bouquets)
4. Subtracting upper probability fraction (something we have to exclude) from 1.

[Reveal] Spoiler:
I don't understand the solution method being used here. I translated the question being asked as "what is the probability that the first two flower bouquet that the florist picked was NOT a bouquet of two same flowers, meaning what is the probability that the first two flower bouquet that the florist picked was either AB OR BP OR AP?". Basically I used probability tree to solve this but I got different answer : 13/36.

AB = One flower is Azalea AND second flower is Buttercup
2/9 x 3/8 = 1/12

OR

BP = one flower is buttercup and second is petunia
3/9 x 4/8 = 1/6

OR

AP = one flower is azalea and second is petunia
2/9 x 4/8 = 1/9

Final answer = 1/12 + 1/6 + 1/9 = 13/36

How is my thinking wrong?

Merging similar tropics. Please refer to the discussion above.

_________________
Intern
Joined: 20 Aug 2015
Posts: 5
Followers: 0

Kudos [?]: 4 [0], given: 83

A florist has 2 azaleas, 3 buttercups, and 4 petunias. She [#permalink]

Show Tags

18 Sep 2015, 03:58
Hi All,
,

This also one way to solve this, it takes maybe slightly longer. This approach is not efficient when dealing with larger "cases".

Azaleas = A
Buttercups = B
Petunias = B

Recap: For the florist to not change the bouquet, different flowers would have been picked.
So, We need to find the probability of three different cases.
P(AB)+P(AP) + P(BP).

P(AB) = 2/9 * 3/8 = 6/72
P(AP) = 2/9 * 4/8 = 8/72
P(BP) = 4/9 * 3/8 = 12/72

Now in each case, we have two different outcomes (AB + BA)

2*(6/72+ 8/72+12/72) = 52/72 = 13/18
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15527
Followers: 651

Kudos [?]: 211 [0], given: 0

Re: A florist has 2 azaleas, 3 buttercups, and 4 petunias. She [#permalink]

Show Tags

10 Apr 2017, 21:36
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Manager
Joined: 13 Dec 2013
Posts: 171
Location: United States
GMAT 1: 710 Q46 V41
GMAT 2: 720 Q48 V40
GPA: 4
WE: Consulting (Consulting)
Followers: 2

Kudos [?]: 19 [0], given: 122

Re: A florist has 2 azaleas, 3 buttercups, and 4 petunias. She [#permalink]

Show Tags

12 Apr 2017, 19:44
ajit257 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?

A. 5/18
B. 13/18
C. 1/9
D. 1/6
E. 2/9

9 flowers. 9C2 = 36 pairs possible.
How many pairs of the same flower? 2C2=1. 3C2=3. 4C2=6. Total of 10 pairs with the same flower. So there are 36-10=26 possible pairs with different flowers. 26/36=13/18.

Kudos if you agree with the method! Please comment if you have improvements.
Re: A florist has 2 azaleas, 3 buttercups, and 4 petunias. She   [#permalink] 12 Apr 2017, 19:44
Similar topics Replies Last post
Similar
Topics:
3 Group A has 2 boys and 3 girls,group B has 3 boys and 4 girls and grou 1 26 Jun 2015, 01:49
12 Grace has 16 jellybeans in her pocket. She has 8 red ones, 4 6 19 Apr 2017, 22:51
14 A bag has 4 blue, 3 yellow and 2 green balls. The balls of 8 03 Mar 2017, 10:36
12 Florist 2azaleas, 3 buttercups and 4 petunias 7 06 Sep 2011, 12:28
9 2^(4-1)^2/2^(3-2) 14 13 Jan 2016, 10:21
Display posts from previous: Sort by