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e12star Just gettin' started! Default Avatar
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Probability Question Post Fri Jan 13, 2012 10:57 am
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  • Lap #[LAPCOUNT] ([LAPTIME])
    Question from Word Translations (Manhattan GMAT), chpt 5:
    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?

    I understand how to find the number of different 2-flower bouquets (by using the anagram model). However, is there a mathematical way to find the number of different bouquets in which both flowers are the same (without actually writing them all out)?

    Thanks!

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    Mike@Magoosh GMAT Instructor
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    Post Fri Jan 13, 2012 12:14 pm
    Hi, there. I'm happy to help with this. Smile

    It sounds like you are already comfortable finding the total number of pairs from a set of nine --- 9C2 =36. Of those 36 pairs, how many would have two of the same flower?

    Well, we have 2 azaleas, 3 buttercups, and 4 petunias. Here we have three different questions to answer:
    a) How many different ways are there to pick 2 azaleas from 2? Just 1
    b) How many different ways are there to pick 2 buttercups from 3? 3C2 = 3
    c) How many different ways are there to pick 2 petunias from 4? 4C2 = 6

    So, there are 1 + 3 + 6 = 10 different pairs that consist of two of the same flower. Probability of two of the same 10/36 = 5/18.

    Does that make sense? Please let me know if you have any questions on what I've said here.

    Mike Smile

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    Brent@GMATPrepNow GMAT Instructor
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    Post Sat Jan 14, 2012 8:53 am
    e12star wrote:
    Question from Word Translations (Manhattan GMAT), chpt 5:
    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?
    I thought I'd mention that we can also solve this question using the rules of probability.

    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

    Similarly: P(both B's) = (3/9)(2/8) = 6/72

    And: 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
    = 13/18

    Cheers,
    Brent

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    MakeUrTimeCount Really wants to Beat The GMAT! Default Avatar
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    Post Sat Jan 14, 2012 9:36 am
    Another Approach:
    We can have different flowers only when we have one of the following 3 combinations:
    1) azaleas, buttercups
    2) azaleas, buttercups
    3) buttercups, petunias.

    For option1), number of ways : 2C1.3C1/9C2 = 6/36
    For option2), number of ways : 2C1.4C1/9C2 = 8/36
    For option3), number of ways : 3C1.4C1/9C2 = 12/36
    Addiing them all: 26/36 = 13/18
    Smile



    Last edited by MakeUrTimeCount on Sat Jan 14, 2012 11:00 am; edited 1 time in total

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    ronnie1985 GMAT Destroyer!
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    Post Sat Jan 14, 2012 10:32 am
    I think there is a serious flaw in the reasoning as now the bouquet is already having 2 flowers of the same color. The florist has to take one of the flower, the color of which is not known and replace it with another flower of some other color. This calls for Baye's Theorem.

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    Brent@GMATPrepNow GMAT Instructor
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    Post Sat Jan 14, 2012 11:03 am
    ronnie1985 wrote:
    I think there is a serious flaw in the reasoning as now the bouquet is already having 2 flowers of the same color. The florist has to take one of the flower, the color of which is not known and replace it with another flower of some other color. This calls for Baye's Theorem.
    We can use Baye's Theorem here. We'll choose the 1st flower and then choose a diff 2nd flower.

    Now, when it comes to choosing the first flower there are 3 cases:

    case 1: choose azalea first, choose different flower second
    The probability of choosing an azalea first is 2/9
    Once we have selected an azalea first, there are 8 flowers remaining (1 A, 3 B's and 4 P's), and we must choose a different flower second.
    So, the probability of choosing a different flower second is 7/8
    So, P(azalea first and different flower second) = (2/9)(7/8) = 14/72

    case 2: choose buttercup first, choose different flower second
    The probability of choosing an buttercup first is 3/9
    Once we have selected an buttercup first, there are 8 flowers remaining (2 A's, 2 B's and 4 P's), and we must choose a different flower second.
    So, the probability of choosing a different flower second is 6/8
    So, P(buttercup first and different flower second) = (3/9)(6/8) = 18/72

    case 3: choose petunia first, choose different flower second
    The probability of choosing an petunia first is 4/9
    Once we have selected an petunia first, there are 8 flowers remaining (2 A's, 3 B's and 3 P's), and we must choose a different flower second.
    So, the probability of choosing a different flower second is 5/8
    So, P(petunia first and different flower second) = (4/9)(5/8) = 20/72


    P(2 diff colors) = (14/72) + (18/72) + (20/72)
    = 52/72
    = 13/18

    Cheers,
    Brent

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    ArunangsuSahu GMAT Destroyer! Default Avatar
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    Post Wed Jan 18, 2012 11:01 pm
    Easily done by

    P(different) = 1-P(same)

    P(same)=(2C2+3C2+4C2)/9c2=5/18

    P(different)=1-5/18
    =13/18

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