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Probability Q From Manhattan

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by sch » Sun May 30, 2010 11:33 pm
Following Questions can be found on Pg 93 of Manhattan GMAT Word Translation Guide

1) A florist has 2 azaleas, 3 buttercups, 4 petunias. She puts two flowers together at random 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 bouguet? If anyone can walk me through this problem step by step very slowly, infact pretend im an actuall idiot, cause I dont even get the explanation provided by the guide on this problem!

2) Obriviated version. 5 actresses: A, B, C, D, E, F to be selected to star in a movie. There are 3 spaces. What are the chances that both A and B will star together. On this one I get o the point where its 1/20 chance, but I dont get why it has to be multiplied by 6. We already know that for the two to be selected together the odds are 1/20 is this now what the problem is asking? The answer is 1/20 (6)=3/10 since there are 6 ways to arrange the outcome when A and B are together.
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by liferocks » Sun May 30, 2010 11:46 pm
Question 1
total number of flowers florist has=2+3+4=9
she can select 2 flowers from these 9 flowers in 9C2=9*8/2=36 ways
customer wants 2 flowers of different types.So
P(Both flowers are of different type)=1-P(Both flowers are of same type)

Same type flowers can be chosen in 3 ways --AA,BB,PP..where A=azaleas,B=buttercups and P=petunias
AA can be chosen from 2 azaleas in 2C2=1 way
BB can be chosen from 3 buttercups in 3C2=3 way
PP can be chosen from 4 petunias in 4C2=6 way
P(Both flowers are of same type)=(1/36)+(3/36)+(6/36)=10/36 or 5/18

so P(Both flowers are of different type)=1-(5/18)=13/18
Last edited by liferocks on Sun May 30, 2010 11:47 pm, edited 1 time in total.
"If you don't know where you are going, any road will get you there."
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by liferocks » Sun May 30, 2010 11:46 pm
Question 2

3 stars can be selected from 5 actresses in 5C3=10 ways
now A and B will always be selected,so only one place is left after selection of A and B .This 1 place can be filled up by any one of D,E or F.So the number of ways this can be done is 3C1=3

hence ans is 3/10
"If you don't know where you are going, any road will get you there."
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by jube » Mon May 31, 2010 2:12 am
I agree with liferock's working (for the florist question) - I did the same and arrived at the same figures but I'm looking at the question & it says: What is the probability that the florist does not have to change the bouguet?

Wouldn't that be 1/2?
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