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

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A florist has 2 azaleas, 3 buttercups, and 4 petunias. She [#permalink] New post 22 Jan 2011, 08:25
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D
E

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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

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Last edited by Bunuel on 17 Nov 2012, 04:36, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Probability ..tough one [#permalink] New post 22 Jan 2011, 08:34
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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}

Answer: B.
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Re: Probability ..tough one [#permalink] New post 05 Feb 2011, 09:21
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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}

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Pls help with explanation for this problem from MGMAT Strategy Guide 5 [#permalink] New post 01 Sep 2014, 23:00
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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
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Re: Probability ..tough one [#permalink] New post 05 Feb 2011, 09:37
##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}
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Florist 2azaleas, 3 buttercups and 4 petunias [#permalink] New post 16 Nov 2012, 19:19
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?
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Pls help with explanation for this problem from MGMAT Strategy Guide 5 [#permalink] New post 01 Sep 2014, 22:28
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?
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Re: A florist has 2 azaleas, 3 buttercups, and 4 petunias. She [#permalink] New post 02 Sep 2014, 02:09
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.

P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention to rules 1,3, 7 and 8. Thank you.


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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: A florist has 2 azaleas, 3 buttercups, and 4 petunias. She   [#permalink] 02 Sep 2014, 02:09
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