Probability1

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Probability1

by gunjan1208 » Wed Oct 05, 2011 12:39 am
Please help me explaning the answer (Source MGMAT strategy guides)

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 sl750 » Wed Oct 05, 2011 12:52 am
P(2 azaleas) = 2/9C2 ; P(2 Buttercup) = 3/9C2; P(2 Petunias) = 4C2/9C2

1 - P(Bouquet with identical flowers) = P(bouquet having 2 different flowers)

2/72+6/72+12/72 = 20/72

1-5/18 = 13/18
Last edited by sl750 on Wed Oct 05, 2011 10:25 am, edited 1 time in total.

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by svd.kumar » Wed Oct 05, 2011 10:11 am
Please post the correct answer, I got a different result. Please correct me if I am wrong

2 azaleas, 3 buttercups, and 4 petunias

Probability of selecting 2 azaleas = 2/9*1/8
Probability of selecting 3 buttercups = 3/9*2/8
Probability of selecting 4 petunias = 4/9*3/8

Which gives 2/72 + 6/72 + 12/72 = 5/18

But we want probability of not having exactly same = 1 - 5/18 = 13/18

Cheers,
SVD

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by sl750 » Wed Oct 05, 2011 10:23 am
I guess I screwed up. 2C2 = 1

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by Anurag@Gurome » Wed Oct 05, 2011 8:03 pm
gunjan1208 wrote:Please help me explaning the answer (Source MGMAT strategy guides)

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?
Let azaleas = A, buttercups = B, and petunias = P

Required probability = 1 - [Probability of both being A + probability of both being B + probability of both being P]
Probability of choosing the same flower = 2/9 * 1/8 + 3/9 * 2/8 + 4/9 * 3/8 = 5/18

Therefore, required probability = 1 = 5/18 = [spoiler]13/18[/spoiler]
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by tpr-becky » Wed Oct 05, 2011 8:05 pm
Probability at its most basic is want over total - so eventually we want to solve for

(bouquet without two same)/ All bouquets

First solve for how many bouquets can you make - with two flowers you have 9 to choose from for the first slot and 8 to choose from in the second slot but order doesn't matter so you divide by 2. Thus 9*8/2 = 36 total bouquets.

Now we need to find the nubmer of bouquets that won't have the same - there ways to look at this
Bouquet has an azalea and another flower (2 x 7)/2 = 7 bouquets
or
Bouquet has a buttercup and another flower: (3 x 6)/2 = 9 bouquets
or
Bouquet has a petunia and another flower: (4 x 5)/2 = 10 bouquets.

add up all bouquets that don't have double flowers - 26 options.

Now put you want over your total - for 26/36 = 13/18 is the final answer
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by gunjan1208 » Thu Oct 06, 2011 2:35 am
Probability of choosing the same flower = 2/9 * 1/8 + 3/9 * 2/8 + 4/9 * 3/8 = 5/18


Why do we multiply with these boldenend numbers?

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by catfreak » Thu Oct 06, 2011 7:57 am
gunjan1208 wrote:Probability of choosing the same flower = 2/9 * 1/8 + 3/9 * 2/8 + 4/9 * 3/8 = 5/18


Why do we multiply with these boldenend numbers?
We are choosing two flower out of 9 one after the other.
So for say buttercup first flower can be chosen in 3/9 and the next buttercup flower after the fist is removed will be chosen in 2/8.

The same logic applies for the rest of the flowers.