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A florist has 2 azaleas, 3 buttercups, and 4 petunias. She [#permalink]
 22 Jan 2011, 09:25
Question Stats:
70% (01:50) correct
29% (00:45) wrong based on 17 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
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Last edited by Bunuel on 17 Nov 2012, 05:36, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Probability ..tough one [#permalink]
22 Jan 2011, 09:34
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Re: Probability ..tough one [#permalink]
05 Feb 2011, 10:21
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]
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}
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Florist 2azaleas, 3 buttercups and 4 petunias [#permalink]
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?
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Florist 2azaleas, 3 buttercups and 4 petunias
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16 Nov 2012, 20:19
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