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Tough one from Kaplan

This topic has 5 expert replies and 2 member replies
aalradadi Junior | Next Rank: 30 Posts Default Avatar
Joined
06 May 2013
Posted:
15 messages

Tough one from Kaplan

Post Tue Dec 10, 2013 5:36 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    A dog breeder currently has 9 breeding dogs. 6 of the dogs have exactly 1 littermate, and 3 of the dogs have exactly 2 littermates. If 2 dogs are selected at random, what is the probability that both selected dogs are NOT littermates?

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    Post Tue Dec 10, 2013 6:00 pm
    Quote:
    A dog breeder has 9 breeding dogs. 6 of the dogs have exactly 1 littermate, and 3 of the dogs have exactly 2 littermates. If 2 dogs are selected at random, what is the probability that both the selected dogs are NOT littermates?
    1/6
    2/9
    5/6
    7/9
    8/9
    Littermates are dogs born in the same batch (litter).

    Here's the counting approach.

    Let the dogs be represented by the letters A to I.
    The following meets the given conditions.
    - A and B are littermates
    - C and D are littermates
    - E and F are littermates
    - G, H and I are littermates

    We want to find P(selected dogs are not littermates)
    Let's use the complement.
    So, P(selected dogs are not littermates) = 1 - P(selected dogs are littermates)

    P(selected dogs are littermates)
    How many possible outcomes satisfy the condition that the two dog ARE littermates?
    We have:
    - A and B
    - C and D
    - E and F
    - G and H
    - G and I
    - H and I
    There are 6 possible outcomes.

    In how many ways can we select 2 dogs from 9 dogs?
    Since order doesn't matter, we can select the dogs in 9C2 ways (36 ways).

    So, P(selected dogs are littermates) = 6/36 = 1/6


    P(selected dogs are not littermates) = 1 - P(selected dogs are littermates)
    = 1 - 1/6
    = 5/6
    = C

    Cheers,
    Brent

    Aside: If anyone is interested, we have a free video on calculating combinations (like 9C2) in your head: http://www.gmatprepnow.com/module/gmat-counting?id=789

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    Post Tue Dec 10, 2013 6:01 pm
    Quote:
    A dog breeder has 9 breeding dogs. 6 of the dogs have exactly 1 littermate, and 3 of the dogs have exactly 2 littermates. If 2 dogs are selected at random, what is the probability that both the selected dogs are NOT littermates?
    1/6
    2/9
    5/6
    7/9
    8/9
    Here's the probability approach.

    Let the dogs be represented by the letters A to I.
    The following meets the given conditions.
    - A and B are littermates
    - C and D are littermates
    - E and F are littermates
    - G, H and I are littermates

    We want to find P(selected dogs are not littermates)

    For the probability approach, it helps to say that one dog is selected first and the other dog is selected second.
    Notice that there are two different ways in which the two dogs are NOT littermates:
    #1) 1st dog is from one of the 2-dog pairings (AB, CD, or EF) and 2nd dog is not a littermate
    #2) 1st dog is from the 3-dog group (GHI) and 2nd dog is not a littermate

    So, . . .
    P(selected dogs are not littermates) = P(1st is from a 2-dog pairing and 2nd is not a littermate OR 1st is from the 3-dog group and 2nd is not a littermate
    = P(1st is from a 2-dog pairing and 2nd is not a littermate) + P(1st is from the 3-dog group and 2nd is not a littermate)
    = (6/9)(7/8) + (3/9)(6/8)
    = 42/72 + 18/72
    = 60/72
    = 5/6
    = C

    Cheers,
    Brent

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    ash4gmat Senior | Next Rank: 100 Posts Default Avatar
    Joined
    17 Sep 2015
    Posted:
    57 messages
    Post Tue May 10, 2016 11:30 pm
    Brent,

    You said Litter mates are born in same batch.Then how come out of 6 dogs you are making 3 pairs when the question tells out of 6 exactly 1 littermate. Same way out of 3 only 2 littermate. Please clarify.

    Post Wed May 11, 2016 2:14 am
    Quote:
    A dog breeder currently has 9 breeding dogs. 6 of the dogs have exactly 1 littermate, and 3 of the dogs have exactly 2 littermates. If 2 dogs are selected at random, what is the probability that both selected dogs are NOT littermates?

    a. 1/6
    b. 2/9
    c. 5/6
    d. 7/9
    e. 8/9
    Let's say that the 9 dogs are ABCDEFGHI.

    6 dogs have exactly 1 littermate:
    Let's say that A and B are littermates, C and D are littermates, and E and F are littermates.
    This means:
    A has 1 littermate (B).
    B has 1 littermate (A).
    C has 1 littermate (D).
    D has 1 littermate (C).
    E has 1 littermate (F).
    F has 1 littermate (E).

    3 dogs have exactly 2 littermates:
    Let's say that G, H and I are all littermates of one another.
    This means:
    G has 2 littermates (H and I).
    H has 2 littermates (G and I).
    I has 2 littermates (G and H).

    Total number of littermate pairs = 6:
    AB, CD, EF, GH, GI, and HI.
    Total number of pairs that can be formed from 9 dogs:
    9C2 = 36.

    P(littermate pair) = 6/36 = 1/6.
    P(not a littermate pair) = 1 - 1/6 = 5/6.

    The correct answer is C.

    If the GMAT were to use the word littermate, a definition would be offered.

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    Post Wed May 11, 2016 8:49 am
    Hi aalradadi,,

    This is a quirky probability question that requires that you keep track of a number of details. There are a few ways to do the math; here's how I would approach it:

    We're told that there are 9 dogs, 6 of them have 1 litter mate and 3 of them have 2 litter mates. ALL of these dogs are contained within the group of 9 dogs.

    So, let's call the dogs:
    1 litter mate:
    A & B
    C & D
    E & F

    2 litter mates:
    G, H and I

    The question asks for the probability that 2 dogs, selected at random, are NOT litter mates.
    I'm going to do the math in 2 calculations:

    If the first dog is one of the "1 litter mate" dogs:
    (6/9)
    then on the next dog, (7/8) are NOT litter mates:
    (6/9)(7/8) = 42/72

    If the first dog is one of the "2 litter mate" dogs:
    (3/9)
    then on the next dog, (6/8) are NOT litter mates:
    (3/9)(6/8) = 18/72

    In TOTAL, (42/72) + (18/72) = 60/72 = 5/6

    Final Answer: C

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    Post Wed May 11, 2016 11:17 pm
    Suppose our dogs are

    AB
    CD
    EF
    GHI

    The first three pairs are obvious. Among GHI, there are three more pairs (GH, GI, HI), for six total.

    Out of the 9 dogs, there are (9 choose 2) = 9*8/2 = 36 pairs.

    So there are 6 littermate pairs out of 36 possible pairs, making it 1/6 that we get littermates and 5/6 that we don't.

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    YTarhouni Newbie | Next Rank: 10 Posts Default Avatar
    Joined
    25 Aug 2017
    Posted:
    8 messages
    Post Sun Sep 03, 2017 2:20 pm
    1-probability no match=1-(3*2/9*1/8+3/9*2/8)=1-(12/72)=5/6

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