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Probability

by jenlee » Fri Mar 29, 2013 4:00 pm
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?
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by srcc25anu » Fri Mar 29, 2013 4:29 pm
# arrangements of any 2 flowers = 9C2 = 36

# arrangements with 2 flowers of the same kind: 2C2 + 3C2 + 4C2 = 10

# arrangements with 2 different flowers = 36 - 10 = 26

Prob that florist does NOT have to make any change post customer call = 26 / 36 = 13/18
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by Brent@GMATPrepNow » Fri Mar 29, 2013 4:44 pm
jenlee 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?
First, we can rewrite the question as "What is the probability that the two flowers are different colors?"

Well, P(diff 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
= [spoiler]13/18[/spoiler]

Cheers,
Brent
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by vipulgoyal » Sun Apr 07, 2013 4:55 am
experts please suggest whats wrong with this approach

2/9 * 3/8 + 2/9 * 4/8 + 3/9 * 4/8 = 13/36
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by Brent@GMATPrepNow » Sun Apr 07, 2013 5:10 am
vipulgoyal wrote:experts please suggest whats wrong with this approach

2/9 * 3/8 + 2/9 * 4/8 + 3/9 * 4/8 = 13/36
It would help if you explained what each value stands for.

Having said that, it looks like you're trying to consider all possible ways in which we can have different flowers. For example, I believe 2/9 * 3/8 stands for the possibility that we select an azalea and then a buttercup. This part is correct P(select azalea then buttercup) = 2/9 * 3/8 = 6/72.

However, your solution also needs to include P(select buttercup then azalea) = 3/9 * 2/8 = 6/72.

In fact, all three parts of your solution are missing the second option.

So, if we double your answer, we get the correct response.

Cheers,
Brent
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by bharat.bondalapati » Sun Apr 07, 2013 5:20 am
You have not accounted for all the cases where flowers of the same kind repeat.

Could you please elaborate on how u arrived at 2/9 * 3/8 + 2/9 * 4/8 + 3/9 * 4/8 = 13/36 ?
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by vipulgoyal » Sun Apr 07, 2013 11:03 pm
Thanks Brent,

I had considered this before, ultimately we are getting one of one kind and second of another so order doesnt matters thats why I put in that way.
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by vipulgoyal » Sun Apr 07, 2013 11:06 pm
I appriciate your help Bharat, The only flaw could be in my way is what the Brent explained, though hard for me to digest
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