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Florist 2azaleas, 3 buttercups and 4 petunias [#permalink]
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16 Mar 2011, 06:34
Question Stats:
57% (03:02) correct 43% (01:43) wrong based on 11 sessions
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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? 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
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Re: Florist 2azaleas, 3 buttercups and 4 petunias [#permalink]
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16 Mar 2011, 06:45
You should multiply each probability by 2, since it could be AzaleaButtercup boquet, or ButtercupAzalea 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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18 Mar 2011, 05:36
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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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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11 Aug 2011, 22:18
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 120/72=52/72=26/36=13/18



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Re: Probability [#permalink]
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11 Aug 2011, 23:35
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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03 Sep 2011, 23:39
Guys, when you use 1p, 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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06 Sep 2011, 12:28
Yangstr wrote: Guys, when you use 1p, 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 110/36 or 26/36 or 13/18



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