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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
00:00
A
B
C
D
E

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Question Stats:

73% (02:18) correct 27% (01:02) wrong based on 33 sessions
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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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?
Florist 2azaleas, 3 buttercups and 4 petunias   [#permalink] 16 Nov 2012, 19:19
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