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

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Difficulty:

25% (medium)

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?

Re: Probability ..tough one [#permalink]
22 Jan 2011, 08:34

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This post received KUDOS

Expert's post

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}

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

gmatclubot

Florist 2azaleas, 3 buttercups and 4 petunias
[#permalink]
16 Nov 2012, 19:19