## Guide 4: Word translations: Chapter 5: Problem 10

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mbolisetty
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### Guide 4: Word translations: Chapter 5: Problem 10

Question: A florist has 2 Azaleas, 3 Buttercups, and 4 Petunias. She put two flowers together at random in a bouquet. However a customer does not want two of the same flower. What is the probability that the florist does not have to change the bouquet ?

I tried answering the above problem in two ways.

1) Prob of having 1 azalea&1 buttercup in a bouquet OR 1 buttercup&1 petunia in a bouque OR 1 Petunia&1 azalea in a bouquet.

=> 2/9*3/8 + 3/9*4/8 + 4/9*2/8 = 26/72

2) 1 - (Prob of having 1 azalea&1azalea OR 1 buttercup&1 buttercup OR 1petunia&1 petunia)

=> 1 - (2/9*1/8 + 3/9*2/8 + 4/9*3/8) = 1 - 20/72 = 13/18

My question is why am I getting TWO different answers. They both have to be the same. 13/18 is the correct answer, as per the Guide. What is the mistake I 'm doing in my FIRST approach
uranus454
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### Re: Guide 4: Word translations: Chapter 5: Problem 10

yes, Even I'm facing the same problem!
Could anyone explain what's wrong with the 1st method??

However, i've found that if you use combinations to solve it, you get the answer as 13/18 in both methods:

Note: I'm using nCr notation = n!/[(n-r)! *r!]

method 1: total cases= 9C2= 36

So we can have A&B "or" B&P "or" P&A
A&B= 2C1*3C1 = 2*3=6
B&P= 3C1*4C1= 3*4=12
P&A= 4C1*2C1= 4*2=8

we add up all these to get 26
now divide by the total number of cases to get 26/36 = 13/18

method 2:

Probability of both flowers different= 1- probability of both flowers same

again total cases = 9C2 = 36

2A "or" 2B "or" 2P

ie 2C2+3C2+4C2 = 1+3+6 = 10

Answer= 1- 10/36 = 26/36 = 13/18

However, i have this doubt:

while computing the total cases as 9C2, why don't we divide by the number of repeats ? ( because there are 2 flowers of 1 kind, 3 flowers of the other kind , 4 flowers of another kind). So, shouldn't we divide it by 2!*3!*4!?

Someone please answer this cuz i have my GMAT on the 13th Feb!!!!!!
Ben Ku
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### Re: Guide 4: Word translations: Chapter 5: Problem 10

mbolisetty wrote:1) Prob of having 1 azalea&1 buttercup in a bouquet OR 1 buttercup&1 petunia in a bouque OR 1 Petunia&1 azalea in a bouquet.

=> 2/9*3/8 + 3/9*4/8 + 4/9*2/8 = 26/72

With this approach, you are setting up a permutation. When you're setting up 1 azalea + 1 buttercup as 2/9 * 3/8, this means you're picking the azalea first and then then buttercup. If order matters, then the TOTAL arrangements are 9*8 or 72.

However, in this problem, order does NOT matter. We don't care whether we pick the azalea first then the buttercup, or the other way. So if we use your approach to set up the probability, it would be:

AB + BA + BP + PB + PA + AP. If you do this, you will get the correct answer.

uranus454 wrote:method 1: total cases= 9C2= 36

This is set up correctly, since you're using a combination to determine the total cases.
Ben Ku
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svyessayan
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### Re: Guide 4: Word translations: Chapter 5: Problem 10

If the equation is (AB + BA + BP + PB + PA + AP) / 36, I end up with 26/18. How do we get to 13/18 because I thought 26/18 was P that flowers will not match....
tim
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### Re: Guide 4: Word translations: Chapter 5: Problem 10

Your numerator appears to be calculated by taking order into account, but the denominator is 9C2 which we get from ignoring order. If we look at all the ordered possibilities, we get 72 rather than 36, because there 9 choices for the first flower * 8 choices for the second flower.
Tim Sanders
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Kovid
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### Re: Guide 4: Word translations: Chapter 5: Problem 10

I have one question regarding the calculation of total number of cases

while computing the total cases as 9C2, why don't we divide by the number of repeats ? ( because there are 2 flowers of 1 kind, 3 flowers of the other kind , 4 flowers of another kind). So, shouldn't we divide it by 2!*3!*4!?

Thanks
RonPurewal
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### Re: Guide 4: Word translations: Chapter 5: Problem 10

Kovid wrote:I have one question regarding the calculation of total number of cases

while computing the total cases as 9C2, why don't we divide by the number of repeats ? ( because there are 2 flowers of 1 kind, 3 flowers of the other kind , 4 flowers of another kind). So, shouldn't we divide it by 2!*3!*4!?

Thanks

^^ ...because then you wouldn't be calculating a probability anymore!

when you make a fraction for probability (desired outcomes divided by total outcomes), the point is that you have to count instances that have the SAME probability.
the "repeats" are actually distinct outcomes with the same probability -- so, a probability calculation is meaningless if you get rid of the "repeats".

__
RonPurewal
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### Re: Guide 4: Word translations: Chapter 5: Problem 10

if this doesn't make sense to you, then consider picking 2 flowers out of, say, 2 roses and 998 daffodils.

if i ask you "what's the probability of picking 2 roses?", then, that's obviously a VERY small number (just by pure common sense).
...but, if you "get rid of the repeats", you're going to mistakenly conclude that the probability is something like 1/3 or 1/4 -- and, even more absurdly, you're going to think it's the same as the probability of getting 2 daffodils (which -- again according to pure common sense -- should be very close to 1, as this is a near certainty).