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

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22 Jan 2011, 09:25

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

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 = \(\frac{72}{2} = 36\)

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.

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.

Answer: B
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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?

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?

Answer:

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

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?

Answer:

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.

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).

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

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10 Apr 2017, 21:36

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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.