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Florist 2azaleas, 3 buttercups and 4 petunias

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

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New post 16 Mar 2011, 06:34
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10) 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 stays that she does not want two of the same flower.What is the probability that the florist does not have to change the bouquet?

Spoiler: :: Peek
Rephrase:- probability that both the two flowers selected are different.

probability that she select one azaleas and one 1 buttercups

2/9*3/8 = 1/12

probability that she select one buttercups and one petunias
3/9*4/8= 1/6

probability that she select one petunias and one azaleas

4/9*2/8 =1/9

total probability that both the flowers ill be different -- 1/12+1/6+1/9 = 13/36

Guys whts wrong in this method


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

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New post 16 Mar 2011, 06:45
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You should multiply each probability by 2, since it could be Azalea-Buttercup boquet, or Buttercup-Azalea boquet, so you have corresponding probabilities 1/6, 1/3, 2/9. Actually 2 is a number of possible combinations of 2 objects. So the answer is 1/6+1/3+2/9=(6+12+8)/36=13/18
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Re: Florist 2azaleas, 3 buttercups and 4 petunias [#permalink]

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New post 18 Mar 2011, 05:36
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Another way to solve it is to consider the circumstances that she WOULD have to change the bouquet, and then subtract that probability that from one.

I found it to be easier since there are fewer combinations to account for:
A1+A2, B1+B2, B1+B3, B2+B3, P1+P2, P1+P3, P1+P4, P2+P3, P2+P4, P3+P4

10/36 is the probability that she WILL have to change the bouquet, so 26/36 or 13/18 is the probability that she will not.
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Probability [#permalink]

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New post 11 Aug 2011, 21:49
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?

5/18
13/18
1/9
1/6
2/9
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Re: Probability [#permalink]

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New post 11 Aug 2011, 22:18
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1
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
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Re: Probability [#permalink]

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New post 11 Aug 2011, 23:35
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required probability = 1- [ probability of both being A'z + probability of both being B'p + probability of both being P'n]

= 1- [ 2/9 * 1/8 + 3/9 * 2/8 + 4/9 * 3/8 ] = 1- 5/18 = 13/18.

Hence B.
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Re: Florist 2azaleas, 3 buttercups and 4 petunias [#permalink]

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New post 03 Sep 2011, 23:39
Guys, when you use 1-p, why don't you take into account the order and multiply the result, in other words there are 2 azaleas so you should imo

2/9*1/8*2 as you can change order of choosing azalea 1 and 2

I understand that it is wrong from the answer but stlill can't get it
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Re: Florist 2azaleas, 3 buttercups and 4 petunias [#permalink]

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New post 06 Sep 2011, 12:28
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1
Yangstr wrote:
Guys, when you use 1-p, why don't you take into account the order and multiply the result, in other words there are 2 azaleas so you should imo

2/9*1/8*2 as you can change order of choosing azalea 1 and 2

I understand that it is wrong from the answer but stlill can't get it



ok.. dint really get your q... can you pls elucidate...

however here is my method... prob(diff flowers) = 1- prob (same flowers)

so prob of having same 2 flowers is . for 2 azaleas can be chosen in 2c2 or 1 way

similarly 2 buttercups can be chosen from 3 in 3c2 or 3 ways

and 2 pertunias from 4 in 4c2 or 6 ways...

therefore total ways is 1+3+6 or 10

total number of possible outcomes is 9c2 or 36

therefore probability of getting the same flowers is 10/36. Probability of not having same flowers is 1-10/36 or 26/36 or 13/18
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Re: Florist 2azaleas, 3 buttercups and 4 petunias [#permalink]

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Re: Florist 2azaleas, 3 buttercups and 4 petunias   [#permalink] 09 Dec 2017, 12:44
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