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A florist has 2 azaleas, 3 buttercups, and 4 petunias. She p [#permalink]
06 Nov 2005, 22:39

00:00

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Difficulty:

35% (medium)

Question Stats:

73% (03:10) correct
27% (01:59) wrong based on 44 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?

gamjatang, does it make a difference if she picks "2 flowers together"?
your explanation seems picking 2 one after other . Just a thought, I am not able to find out, the different way though.

Re: A florist has 2 azaleas, 3 buttercups, and 4 petunias. She [#permalink]
23 Mar 2014, 11:00

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Re: A florist has 2 azaleas, 3 buttercups, and 4 petunias. She p [#permalink]
24 Mar 2014, 00:30

Expert's post

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

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}