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

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Joined: 04 Nov 2005
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A florist has 2 azaleas, 3 buttercups, and 4 petunias. She p [#permalink]

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06 Nov 2005, 23:39
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Question Stats:

67% (02:09) correct 33% (02:31) wrong based on 198 sessions

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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

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-florist-has-2-azaleas-3-buttercups-and-4-petunias-she-108116.html
[Reveal] Spoiler: OA

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Director
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06 Nov 2005, 23:51
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Total flowers : AA, BBB, PPPP

Probability that two same flowers are picked:

AA = 2/9 * 1/8 = 1/36
BB = 3/9 * 2/8 = 1/12
PP = 4/9 * 3/8 = 1/6
--------------------------
1/36 + 1/12 + 1/6 = 5/18

1 - 5/18 = 13/18

Thus, 13/18
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06 Nov 2005, 23:55
gamjatang wrote:
Total flowers : AA, BBB, PPPP

Probability that two same flowers are picked:

AA = 2/9 * 1/8 = 1/36
BB = 3/9 * 2/8 = 1/12
PP = 4/9 * 3/8 = 1/6
--------------------------
1/36 + 1/12 + 1/6 = 5/18

1 - 5/18 = 13/18

Thus, 13/18

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.

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07 Nov 2005, 00:03
gamjatang wrote:
Total flowers : AA, BBB, PPPP

Probability that two same flowers are picked:

AA = 2/9 * 1/8 = 1/36
BB = 3/9 * 2/8 = 1/12
PP = 4/9 * 3/8 = 1/6
--------------------------
1/36 + 1/12 + 1/6 = 5/18

1 - 5/18 = 13/18

Thus, 13/18

Thanks for the help. 13/18 is the answer in the book but I cant see how to get to it.

Where do you get the "*1/8" "*2/8" "*3/8"

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Director
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07 Nov 2005, 02:23
cjrylant wrote:
gamjatang wrote:
Total flowers : AA, BBB, PPPP

Probability that two same flowers are picked:

AA = 2/9 * 1/8 = 1/36
BB = 3/9 * 2/8 = 1/12
PP = 4/9 * 3/8 = 1/6
--------------------------
1/36 + 1/12 + 1/6 = 5/18

1 - 5/18 = 13/18

Thus, 13/18

Thanks for the help. 13/18 is the answer in the book but I cant see how to get to it.

Where do you get the "*1/8" "*2/8" "*3/8"

// cjrylant

There are 9 flowers now.

After you pick one, then there are 8 flowers left.

Likewise, there are 4 PPPP now.

After you pick one P, then there are three PPP left.

That's where "1/8", "2/8", and "3/8" came from.

// duttsit

If you pick two flowers at one time, the result will be the same.
The formula will be like following (and is much more time-consuming);

1) Total number of possible picks
= 9C2 = 9*8/2=36

2) Possible number of cases in which two A's are picked
= 2C2 = 1

2) Possible number of cases in which two B's are picked
= 3C2 = 3

3) Possible number of cases in which two P's are picked
= 4C2 = 6

-----------------------------------------------------------------

From 1), 2), 3), and 4), we can see that the possibility of picking two same flowers is (1+3+6)/36 = 10/36 = 5/18

Thus, the possibility of picking two different flowers is 1 - 5/18 = 13/18.
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Re: A florist has 2 azaleas, 3 buttercups, and 4 petunias. She [#permalink]

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23 Mar 2014, 12:00
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Re: A florist has 2 azaleas, 3 buttercups, and 4 petunias. She p [#permalink]

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24 Mar 2014, 01:30
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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

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: $$\frac{C^2_2+C^2_3+C^2_4}{C^2_{9}}=\frac{10}{36}$$ --> $$P=1-\frac{10}{36}=\frac{26}{36}=\frac{13}{18}$$

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-florist-has-2-azaleas-3-buttercups-and-4-petunias-she-108116.html
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Re: A florist has 2 azaleas, 3 buttercups, and 4 petunias. She p   [#permalink] 24 Mar 2014, 01:30
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