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florist (probability)

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florist (probability)

by rahul.s » Mon Feb 15, 2010 11:09 pm
a florist has 2 roses, 3 buttercups, and 4 orchids. she puts 2 flowers together at random in a bouquet. however, the customer calls and says that she does not want 2 of the same flower. what is the probability that the florist does not have to change the bouquet?

sorry, but there aren't any answer choices.

OA: [spoiler]13/18[/spoiler]
Source: mgmat
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by ajith » Mon Feb 15, 2010 11:41 pm
rahul.s wrote:a florist has 2 roses, 3 buttercups, and 4 orchids. she puts 2 flowers together at random in a bouquet. however, the customer calls and says that she does not want 2 of the same flower. what is the probability that the florist does not have to change the bouquet?

sorry, but there aren't any answer choices.

OA: [spoiler]13/18[/spoiler]
Source: mgmat
No of ways in which one can select two flowers from 9 (3+4+2) = 9C2 = 36
No of ways in which one can select two flowers of the same kind = 2C2 + 3C2+ 4C2 = 1+3+6 =10
Probability that the two flowers that selected that of same kind = 10/36 = 5/18
Probability that the florist does not have to change the bouquet = 1 -Probability that the two flowers that selected that of same kind = 1-5/18 = 13/18
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by thephoenix » Tue Feb 16, 2010 12:19 am
tot no. of wyas for selecting 2 flower out of 9 flower=9C2=36

lets find out the cases where the two flowers are same
P of two flowers to be roses=2C2/9C2=1/36
P of two flowers to be buttecups=3C2/9C2=3/36
P of two flowers to be buttecups=4C2/9C2=6/36

tot P=add all=10/36=5/18

reqd P=1-tot P for selceting two of same kind
=13/18
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by kvamsy » Tue Feb 16, 2010 5:26 am
rahul.s wrote:a florist has 2 roses, 3 buttercups, and 4 orchids. she puts 2 flowers together at random in a bouquet. however, the customer calls and says that she does not want 2 of the same flower. what is the probability that the florist does not have to change the bouquet?

sorry, but there aren't any answer choices.

OA: [spoiler]13/18[/spoiler]
Source: mgmat
One more way to solve.

Total number of ways of selecting 2 flowers = 9C2 = 36
2 flowers should select at random = (2C1*3C1)+(2C1*4C1)+(3C1*4C1)

probability = 26/36 -----> 13/26
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by rahul.s » Tue Feb 16, 2010 5:34 am
kvamsy wrote:One more way to solve.

Total number of ways of selecting 2 flowers = 9C2 = 36
2 flowers should select at random = (2C1*3C1)+(2C1*4C1)+(3C1*4C1)

probability = 26/36 -----> 13/26
i think you meant 13/18 :)
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by kvamsy » Tue Feb 16, 2010 5:51 am
rahul.s wrote:
kvamsy wrote:One more way to solve.

Total number of ways of selecting 2 flowers = 9C2 = 36
2 flowers should select at random = (2C1*3C1)+(2C1*4C1)+(3C1*4C1)

probability = 26/36 -----> 13/26
i think you meant 13/18 :)
you are right
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