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Math Help - Two dice

  1. #1
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    Two dice

    If you had two six sided dice and you tossed them, what would be the probability that the sum is 5 or less?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by gretchen View Post
    If you had two six sided dice and you tossed them, what would be the probability that the sum is 5 or less?
    i'm quite sure there's a neat little formula for this, but i don't remember it, so let's do it the long and hard way.

    we want the sum to be at least 5, so it can be 5,4,3 or 2

    how many ways are there to get a sum of 5?
    1) Dice1 = 4 AND Dice2 = 1, Probability = (1/6)*(1/6) = 1/36
    2) Dice1 = 3 AND Dice2 = 2, Probability = (1/6)*(1/6) = 1/36
    3) Dice1 = 2 AND Dice2 = 3, Probability = (1/6)*(1/6) = 1/36
    4) Dice1 = 1 AND Dice2 = 4, Probability = (1/6)*(1/6) = 1/36

    so probability of one of these ways happening is: 1/36 + 1/36 + 1/36 + 1/36 = 4/36 = 1/9

    how many ways are there to get a sum of 4?
    1) Dice1 = 3 AND Dice2 = 1, Probability = (1/6)*(1/6) = 1/36
    2) Dice1 = 2 AND Dice2 = 2, Probability = (1/6)*(1/6) = 1/36
    3) Dice1 = 1 AND Dice2 = 3, Probability = (1/6)*(1/6) = 1/36

    so probability of one of these ways happening is: 1/36 + 1/36 + 1/36 = 3/36 = 1/12

    how many ways are there to get a sum of 3?
    1) Dice1 = 2 AND Dice2 = 1, Probability = (1/6)*(1/6) = 1/36
    2) Dice1 = 1 AND Dice2 = 2, Probability = (1/6)*(1/6) = 1/36

    so probability of one of these ways happening is: 1/36 + 1/36 = 2/36 = 1/18

    how many ways are there to get a sum of 2?
    1) Dice1 = 1 AND Dice2 = 1, Probability = (1/6)*(1/6) = 1/36

    so probability of one of these ways happening is 1/36

    So the probability of getting a sum of 5 OR 4 OR 3 OR 2 is:

    1/9 + 1/12 + 1/18 + 1/36 = 5/18

    so, the probability that the sum is 5 or less is 5/18
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  3. #3
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    Hello, Gretchen!

    If you had two six-sided dice and you tossed them,
    what would be the probability that the sum is 5 or less?

    If you're just starting Probability, you might get used to dice problems.

    With two dice, there are 36 possible outcomes.

    (1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
    (2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
    (3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
    (4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
    (5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
    (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)


    You don't have to write them out every time.
    Now that you've seen them, you can visualize them, right?

    Now pick out and count the ones you want.
    "Five or less": (1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)
    There are ten of them.

    Therefore: P(5 or less) .= .10/36 .= .5/18

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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, Gretchen!


    If you're just starting Probability, you might get used to dice problems.

    With two dice, there are 36 possible outcomes.

    (1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
    (2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
    (3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
    (4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
    (5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
    (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)


    You don't have to write them out every time.
    Now that you've seen them, you can visualize them, right?

    Now pick out and count the ones you want.
    "Five or less": (1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)
    There are ten of them.

    Therefore: P(5 or less) .= .10/36 .= .5/18

    Ah, yes, i forgot you could set up such a table. I did probability ages ago, i'm a bit rusty on it as i hinted in my first post
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