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Bill

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j_shreyans GMAT Destroyer! Default Avatar
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Bill Post Wed Oct 29, 2014 8:12 am
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  • Lap #[LAPCOUNT] ([LAPTIME])
    Bill has a set of 6 black cards and a set of 6 red cards. Each card has a number from 1 through 6, such that each of the numbers 1 through 6 appears on 1 black card and 1 red card. Bill likes to play a game in which he shuffles all 12 cards, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?

    A)8/33

    B)62/165

    C)17/33

    D)103/165

    E)25/33

    OAC

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    Post Wed Oct 29, 2014 9:13 am
    j_shreyans wrote:
    Bill has a set of 6 black cards and a set of 6 red cards. Each card has a number from 1 through 6, such that each of the numbers 1 through 6 appears on 1 black card and 1 red card. Bill likes to play a game in which he shuffles all 12 cards, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?

    A)8/33

    B)62/165

    C)17/33

    D)103/165

    E)25/33
    P(at least 1 pair) = 1 - P(no pairs).

    P(no pairs):
    Any of the 12 cards can be selected first.
    P(2nd card does not match the first) = 10/11. (Of the 11 remaining cards, any but the mate of the 1st card.)
    P(3rd card does not match the first 2 cards) = 8/10. (Of the 10 remaining cards, any but the mates of the first 2 cards.)
    P(4th card does not match the first 3 cards) = 6/9. (Of the 9 remaining cards, any but the mates of the first 3 cards.)
    Since we want all of these events to happen, we MULTIPLY the fractions:
    10/11 * 8/10 * 6/9 = 16/33.

    Thus:
    P(at least 1 pair) = 1 - 16/33 = 17/33.

    The correct answer is C.

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    Post Wed Oct 29, 2014 9:37 am
    j_shreyans wrote:
    Bill has a set of 6 black cards and a set of 6 red cards. Each card has a number from 1 through 6, such that each of the numbers 1 through 6 appears on 1 black card and 1 red card. Bill likes to play a game in which he shuffles all 12 cards, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?

    A)8/33
    B)62/165
    C)17/33
    D)103/165
    E)25/33

    We can solve this question using probability rules.

    First, recognize that P(at least one pair) = 1 - P(no pairs)

    P(no pairs) = P(select any 1st card AND select any non-matching card 2nd AND select any non-matching card 3rd AND select any non-matching card 4th)
    = P(select any 1st card) x P(select any non-matching card 2nd) x P(select any non-matching card 3rd) x P(select any non-matching card 4th)
    = 1 x 10/11 x 8/10 x 6/9
    = 16/33

    So, P(at least one pair) = 1 - 16/33
    = 17/33
    = C

    Cheers,
    Brent

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    Post Wed Oct 29, 2014 9:38 am
    Quote:
    Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, there are 2 cards in the deck that have the same value. Bill likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?
    A. 8/33
    B. 62/165
    C. 17/33
    D. 103/165
    E. 25/33
    We can also solve the question using counting techniques.
    First recognize that P(at least one pair) = 1 - P(no pairs)

    We'll find P(no pairs)

    First, the number of possible outcomes.
    We have 12 cards and we select 4.
    This is accomplished in 12C4 ways = 495

    Aside: If anyone is interested, we have a free video on calculating combinations (like 12C4) in your head: http://www.gmatprepnow.com/module/gmat-counting?id=789

    Now we count the number of ways to select 4 different cards with no pairs. In other words, we want 4 different card values.
    First look at how many different cards we can select (WITHOUT considering the suits)
    There are 6 card values and we want to select 4.
    This is accomplished in 6C4 = 15 ways.
    For each of these 15 values (e.g., 1,2,4,5) each card can be of either suit.
    So, for example, in how many ways can we have card values of 1,2,4,5?
    For the 1-card we have 2 suits from which to choose.
    For the 2-card we have 2 suits from which to choose.
    For the 4-card we have 2 suits from which to choose.
    For the 5-card we have 2 suits from which to choose.
    So, for the selection 1,2,4,5 there are 2x2x2x2=16 possibilities.

    So, the number of ways to select 4 cards such that there are no pairs is 15x16=240

    So, the probability that there are no pairs = 240/495 = 16/33

    So, P(at least one pair) = 1- 16/33 = 17/33 = C

    Cheers,
    Brent

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    Post Fri Oct 31, 2014 9:52 am
    To piggyback on Brent's method:

    Probability = (# of target outcomes)/(# of total outcomes)

    Target outcome = No pairs
    Total outcome = Any group of four cards

    The total is easy: 12 cards choose 4, or 12!/(8! * 4!), or (12*11*10*9)/(4*3*2), or 11*9*5, or 495.

    The target is trickier. First, we need to choose four different numbers of the six. We can do this in (6 choose 4) or 6!/(4! * 2!) or 6*5/2 or 15 ways.

    Next, each of those four numbers could be either black or red, so we have TWO options for each. This gives us our 15 ways (from the above) * 2 * 2 * 2 * 2 (two choices for each card), or 240 ways total.

    Hence target = 240, total = 495, and the probability = 240/495 = 16/33.

    Since this is the probability of NO PAIRS, we know the probability is one pair is 1 - 16/33, or 17/33.

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