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Probability of receiving a five or a one on either die,

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gmattesttaker2 GMAT Destroyer! Default Avatar
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Probability of receiving a five or a one on either die, Post Sun May 04, 2014 8:32 pm
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  • Lap #[LAPCOUNT] ([LAPTIME])
    Hello,

    Can you please assist with this:

    In a game, one player throws two fair, six-sided die at the same time. If the player receives a five or a one on either die, that player wins. What is the probability that a player wins after playing the game once?


    OA: 5/9


    Thanks a lot,
    Sri

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    Post Sun May 04, 2014 10:07 pm
    gmattesttaker2 wrote:
    In a game, one player throws two fair, six-sided die at the same time. If the player receives a five or a one on either die, that player wins. What is the probability that a player wins after playing the game once?
    So, the player wins if he/she rolls AT LEAST one 5 or 1.
    When it comes to probability questions involving "at least," it's best to try using the complement.
    That is, P(Event A happening) = 1 - P(Event A not happening)
    So, here we get: P(AT LEAST one 5 or 1) = 1 - P(zero 5's or 1's)
    = 1 - P(no 5 or 1 on 1st die AND no 5 or 1 on 2nd die)
    = 1 - [P(no 5 or 1 on 1st die) x P(no 5 or 1 on 2nd die)]
    = 1 - [ 4/6 x 4/6]
    = 1 - [16/36]
    = 20/36
    = 5/9

    Cheers,
    Brent

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    gmattesttaker2 GMAT Destroyer! Default Avatar
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    Post Mon May 05, 2014 6:00 pm
    Brent@GMATPrepNow wrote:
    gmattesttaker2 wrote:
    In a game, one player throws two fair, six-sided die at the same time. If the player receives a five or a one on either die, that player wins. What is the probability that a player wins after playing the game once?
    So, the player wins if he/she rolls AT LEAST one 5 or 1.
    When it comes to probability questions involving "at least," it's best to try using the complement.
    That is, P(Event A happening) = 1 - P(Event A not happening)
    So, here we get: P(AT LEAST one 5 or 1) = 1 - P(zero 5's or 1's)
    = 1 - P(no 5 or 1 on 1st die AND no 5 or 1 on 2nd die)
    = 1 - [P(no 5 or 1 on 1st die) x P(no 5 or 1 on 2nd die)]
    = 1 - [ 4/6 x 4/6]
    = 1 - [16/36]
    = 20/36
    = 5/9

    Cheers,
    Brent
    Hello Brent,

    Thank you very much for your excellent and detailed explanation.

    Best Regards,
    Sri

    Post Mon May 05, 2014 10:43 pm
    Hi Sri,

    Brent's solution is great (and it's exactly how I would approach the question). There is another way to approach it though. Since you're rolling 2 dice, there aren't that many possible outcomes (just 36 in total), so you COULD use brute force and just write them all down:

    We're looking for the number of outcomes that include AT LEAST a 1 or a 5.

    1,1
    1,2
    1,3
    1,4
    1,5
    1,6

    2,1
    2,5

    3,1
    3,5

    4,1
    4,5

    5,1
    5,2
    5,3
    5,4
    5,5
    5,6

    6,1
    6,5

    Total possibilities = 20

    Probability of rolling at least a 1 or a 5 on two dice: 20/36 = 5/9

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    Rich

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    Post Tue May 06, 2014 7:04 pm
    Rich.C@EMPOWERgmat.com wrote:
    Hi Sri,

    Brent's solution is great (and it's exactly how I would approach the question). There is another way to approach it though. Since you're rolling 2 dice, there aren't that many possible outcomes (just 36 in total), so you COULD use brute force and just write them all down:

    We're looking for the number of outcomes that include AT LEAST a 1 or a 5.

    1,1
    1,2
    1,3
    1,4
    1,5
    1,6

    2,1
    2,5

    3,1
    3,5

    4,1
    4,5

    5,1
    5,2
    5,3
    5,4
    5,5
    5,6

    6,1
    6,5

    Total possibilities = 20

    Probability of rolling at least a 1 or a 5 on two dice: 20/36 = 5/9

    GMAT assassins aren't born, they're made,
    Rich
    Hello Rich,

    Thanks a lot for the alternate approach.

    Best Regards,
    Sri

    gmattesttaker2 GMAT Destroyer! Default Avatar
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    Post Wed May 07, 2014 8:03 am
    Brent@GMATPrepNow wrote:
    gmattesttaker2 wrote:
    In a game, one player throws two fair, six-sided die at the same time. If the player receives a five or a one on either die, that player wins. What is the probability that a player wins after playing the game once?
    So, the player wins if he/she rolls AT LEAST one 5 or 1.
    When it comes to probability questions involving "at least," it's best to try using the complement.
    That is, P(Event A happening) = 1 - P(Event A not happening)
    So, here we get: P(AT LEAST one 5 or 1) = 1 - P(zero 5's or 1's)
    = 1 - P(no 5 or 1 on 1st die AND no 5 or 1 on 2nd die)
    = 1 - [P(no 5 or 1 on 1st die) x P(no 5 or 1 on 2nd die)]
    = 1 - [ 4/6 x 4/6]
    = 1 - [16/36]
    = 20/36
    = 5/9

    Cheers,
    Brent
    Hello Brent,

    I was just wondering if we can apply the same technique of

    P(Event A happening) = 1 - P(Event A not happening)

    for the following as well or should this technique be strictly applied for only the ones that have "at least".


    Mathematics, physics, and chemistry books are stored on a library shelf that can accommodate 25 books. Currently, 20% of the shelf spots remain empty. There are twice as many mathematics books as physics books and the number of physics books is 4 greater than that of the chemistry books. Among all the books, 12 books are softcover and the remaining are hard-cover. If there are a total of 7 hard-cover books among the mathematics and physics books, what is the probability that a book selected at random is either a hard-cover book or a chemistry book?

    OA: 9/20

    m + p + c = 80/100(25)
    => m + p + c = 20

    m = 2p and p = 4 + c
    => 2p + p + c = 20
    => 3p + c = 20
    => 3(4 + c) + c = 20
    => 12 + 3c + c = 20
    => 12 + 4c = 20
    => 4c = 8
    => c = 2
    => m + p = 18

    Total Hard covers = 8
    mathematics Hard cover + physics Hard cover = 7 => chemistry Hard cover = 1
    Also, this means Total Soft Covers = 12


    The probability that a book selected at random is either a hard-cover book or a chemistry book
    = 1 - P (Not selecting a book that is either hard cover or chemistry)

    However, I was not very sure how to solve from this point onwards.

    Thanks a lot for your help.

    Best Regards,
    Sri

    e30sport Just gettin' started! Default Avatar
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    Post Fri Jun 12, 2015 8:07 am
    Brent@GMATPrepNow wrote:
    gmattesttaker2 wrote:
    In a game, one player throws two fair, six-sided die at the same time. If the player receives a five or a one on either die, that player wins. What is the probability that a player wins after playing the game once?
    So, the player wins if he/she rolls AT LEAST one 5 or 1.
    When it comes to probability questions involving "at least," it's best to try using the complement.
    That is, P(Event A happening) = 1 - P(Event A not happening)
    So, here we get: P(AT LEAST one 5 or 1) = 1 - P(zero 5's or 1's)
    = 1 - P(no 5 or 1 on 1st die AND no 5 or 1 on 2nd die)
    = 1 - [P(no 5 or 1 on 1st die) x P(no 5 or 1 on 2nd die)]
    = 1 - [ 4/6 x 4/6]
    = 1 - [16/36]
    = 20/36
    = 5/9

    Cheers,
    Brent
    Brent,

    Please explain to me why it's not just P(A and B) = P(A) x P(B). You can have two outcomes on the first and two on the second which would be 2/6 * 2/6.

    I clearly don't understand the principles.

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    Post Fri Jun 12, 2015 11:59 am
    p(1,1) = 1/6 X 1/6 = 1/36
    p(1,2) = 1/6 X 1/6 = 1/36
    p(1,3) = 1/6 X 1/6 = 1/36
    p(1,4) = 1/6 X 1/6 = 1/36
    p(1,5) = 1/6 X 1/6 = 1/36
    p(1,6) = 1/6 X 1/6 = 1/36

    p(2,1) = 1/6 X 1/6 = 1/36
    p(2,5) = 1/6 X 1/6 = 1/36
    p(3,1) = 1/6 X 1/6 = 1/36
    p(3,5) = 1/6 X 1/6 = 1/36
    p(4,1) = 1/6 X 1/6 = 1/36
    p(4,5) = 1/6 X 1/6 = 1/36

    p(5,1) = 1/6 X 1/6 = 1/36
    p(5,2) = 1/6 X 1/6 = 1/36
    p(5,3) = 1/6 X 1/6 = 1/36
    p(5,4) = 1/6 X 1/6 = 1/36
    p(5,5) = 1/6 X 1/6 = 1/36
    p(5,6) = 1/6 X 1/6 = 1/36
    p(6,1) = 1/6 X 1/6 = 1/36
    p(6,5) = 1/6 X 1/6 = 1/36

    Therefore, the total sum is 20/36. So, there is more than two outcomes for the first and second, 2/6 does not account for all the possibilities.



    Last edited by talaangoshtari on Fri Jun 12, 2015 1:00 pm; edited 1 time in total

    Post Fri Jun 12, 2015 12:50 pm
    e30sport wrote:
    Brent,

    Please explain to me why it's not just P(A and B) = P(A) x P(B). You can have two outcomes on the first and two on the second which would be 2/6 * 2/6.

    I clearly don't understand the principles.
    Hi e30sport,

    Applying P(A and B) means finding the probability that we get 5 or 1 on the first die AND 5 or 1 on the second die.
    To win the game we don't need to get 5 or 1 on BOTH dice.

    Cheers,
    Brent

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    Anantjit Just gettin' started! Default Avatar
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    Post Wed Jul 27, 2016 5:55 am
    It can be like below :-
    Probability of 5 or 1 on either dice= P(5) or P(1) on 1st dice OR P(5) or P(1) on 2nd Dice
    i.e P[A]+P[B]-P[A and B]
    P[A]=P[5]+P[1]-P[5 and 1]=1/6+1/6-0=1/3
    P[B]=P[5]+P[1]-P[5 and 1]=1/6+1/6-0=1/3
    P[A and B]= 1/3+1/3-[1/3*1/3]=2/3-1/9=5/9

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    Post Thu Aug 04, 2016 9:16 pm
    Anantjit wrote:
    It can be like below :-
    Probability of 5 or 1 on either dice= P(5) or P(1) on 1st dice OR P(5) or P(1) on 2nd Dice
    i.e P[A]+P[B]-P[A and B]
    P[A]=P[5]+P[1]-P[5 and 1]=1/6+1/6-0=1/3
    P[B]=P[5]+P[1]-P[5 and 1]=1/6+1/6-0=1/3
    P[A and B]= 1/3+1/3-[1/3*1/3]=2/3-1/9=5/9
    This is the right idea, but you want to minimize the amount of clutter on your page during the test, or it's easy to confuse yourself with too much superfluous jargon. (It also probably wouldn't help someone who doesn't understand the idea in the first place, as it's a bit of a wall of text: keeping it simple and clean is good for the teacher and the student Very Happy)

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