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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]

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Updated on: 17 Nov 2012, 05:36
8
00:00

Difficulty:

55% (hard)

Question Stats:

66% (02:17) correct 34% (02:27) wrong based on 232 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

Originally posted by ajit257 on 22 Jan 2011, 09:25.
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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Posts: 49892

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22 Jan 2011, 09:34
2
4
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: $$\frac{C^2_2+C^2_3+C^2_4}{C^2_{9}}=\frac{10}{36}$$ --> $$P=1-\frac{10}{36}=\frac{26}{36}=\frac{13}{18}$$

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05 Feb 2011, 10:21
1
1

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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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]

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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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Pls help with explanation for this problem from MGMAT Strategy Guide 5  [#permalink]

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02 Sep 2014, 00:00
1
What we have: 2A, 3B, 4P
Total: 9 flowers

Prob of selecting 2A: (2/9)(1/8)=1/36
Prob of selecting 2B: (3/9)(2/8)=1/12=3/36
Prob of selecting 2P: (4/9)(3/8)=1/6=6/36

1/36+3/36+6/36=10/36=5/18

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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02 Sep 2014, 03:09
1
arpshriv wrote:
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?

Merging similar tropics. please refer to the discussion above.

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A florist has 2 azaleas, 3 buttercups, and 4 petunias. She puts...  [#permalink]

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02 Sep 2015, 09:49
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?

1. Counting manually (by drawing a table) all possible combinations of bouquets with two same flowes
2. Counting all possible combinations of two flower bouquets by using anagram grid
3. Making probability fraction of amount of possible bouquets with 2 same flowers to all possible bouquets (two same flower and two different flower bouquets)
4. Subtracting upper probability fraction (something we have to exclude) from 1.

I don't understand the solution method being used here. I translated the question being asked as "what is the probability that the first two flower bouquet that the florist picked was NOT a bouquet of two same flowers, meaning what is the probability that the first two flower bouquet that the florist picked was either AB OR BP OR AP?". Basically I used probability tree to solve this but I got different answer : 13/36.

AB = One flower is Azalea AND second flower is Buttercup
2/9 x 3/8 = 1/12

OR

BP = one flower is buttercup and second is petunia
3/9 x 4/8 = 1/6

OR

AP = one flower is azalea and second is petunia
2/9 x 4/8 = 1/9

Final answer = 1/12 + 1/6 + 1/9 = 13/36

How is my thinking wrong?
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Joined: 02 Sep 2009
Posts: 49892
Re: A florist has 2 azaleas, 3 buttercups, and 4 petunias. She  [#permalink]

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02 Sep 2015, 09:56
MariaVorop wrote:
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?

1. Counting manually (by drawing a table) all possible combinations of bouquets with two same flowes
2. Counting all possible combinations of two flower bouquets by using anagram grid
3. Making probability fraction of amount of possible bouquets with 2 same flowers to all possible bouquets (two same flower and two different flower bouquets)
4. Subtracting upper probability fraction (something we have to exclude) from 1.

I don't understand the solution method being used here. I translated the question being asked as "what is the probability that the first two flower bouquet that the florist picked was NOT a bouquet of two same flowers, meaning what is the probability that the first two flower bouquet that the florist picked was either AB OR BP OR AP?". Basically I used probability tree to solve this but I got different answer : 13/36.

AB = One flower is Azalea AND second flower is Buttercup
2/9 x 3/8 = 1/12

OR

BP = one flower is buttercup and second is petunia
3/9 x 4/8 = 1/6

OR

AP = one flower is azalea and second is petunia
2/9 x 4/8 = 1/9

Final answer = 1/12 + 1/6 + 1/9 = 13/36

How is my thinking wrong?

Merging similar tropics. Please refer to the discussion above.

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A florist has 2 azaleas, 3 buttercups, and 4 petunias. She  [#permalink]

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18 Sep 2015, 03:58
Hi All,
,

This also one way to solve this, it takes maybe slightly longer. This approach is not efficient when dealing with larger "cases".

Azaleas = A
Buttercups = B
Petunias = B

Recap: For the florist to not change the bouquet, different flowers would have been picked.
So, We need to find the probability of three different cases.
P(AB)+P(AP) + P(BP).

P(AB) = 2/9 * 3/8 = 6/72
P(AP) = 2/9 * 4/8 = 8/72
P(BP) = 4/9 * 3/8 = 12/72

Now in each case, we have two different outcomes (AB + BA)

2*(6/72+ 8/72+12/72) = 52/72 = 13/18
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Re: A florist has 2 azaleas, 3 buttercups, and 4 petunias. She  [#permalink]

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12 Apr 2017, 19:44
ajit257 wrote:
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

9 flowers. 9C2 = 36 pairs possible.
How many pairs of the same flower? 2C2=1. 3C2=3. 4C2=6. Total of 10 pairs with the same flower. So there are 36-10=26 possible pairs with different flowers. 26/36=13/18.

Kudos if you agree with the method! Please comment if you have improvements.
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Re: A florist has 2 azaleas, 3 buttercups, and 4 petunias. She  [#permalink]

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18 Apr 2017, 16:44
1
ajit257 wrote:
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

We can use the following formula:

1 = P(getting two of the same flower) + P(not getting two of the same flower)

Let’s determine the probability of getting two of the same flower.

P(2 azaleas) = 2/9 x 1/8 = 2/72

P(2 buttercups) = 3/9 x 2/8 = 6/72

P(4 petunias) = 4/9 x 3/8 = 12/72

The probability of getting two of the same flower is, therefore, 2/72 + 6/72 + 12/72 = 20/72 = 5/18

Thus, the probability of not getting two of the same flower 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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24 Apr 2018, 10:54
Top Contributor
ajit257 wrote:
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

First, we can rewrite the question as "What is the probability that the two flowers are different colors?"

Well, P(different colors) = 1 - P(same color)

Aside: let A = azalea, let B = buttercup, let P = petunia

P(same color) = P(both A's OR both B's OR both P's)
= P(both A's) + P(both B's) + P(both P's)

Now let's examine each probability:

P(both A's):
We need the 1st flower to be an azalea and the 2nd flower to be an azalea
So, P(both A's) = (2/9)(1/8) = 2/72

P(both B's)
= (3/9)(2/8) = 6/72

P(both P's)
= (4/9)(3/8) = 12/72

So, P(same color) = (2/72) + (6/72) + (12/72) = 20/72 = 5/18

Now back to the beginning:
P(diff colors) = 1 - P(same color)
= 1 - 5/18
= 13/18

Cheers,
Brent
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Re: A florist has 2 azaleas, 3 buttercups, and 4 petunias. She &nbs [#permalink] 24 Apr 2018, 10:54
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