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Math Help - probability of rolling a die...

  1. #1
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    probability of rolling a die...

    ok this problem is a little different from the others...

    what is the probability of either a die roll of a six or flipping a coin once and getting a heads?

    ok so this means i will use both laws together? first multiply and then add?

    so the probability of a die roll is (1/6) times the probability of getting a heads (1/2)

    so this is 1/12 + (1/6)(1/2)=1/6?

    did i do this right?
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  2. #2
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    You must first decide what "or" means. Does it include both or is it exclusive?

    p(6) = d
    p(not six) = 1-d = f

    p(heads) = h
    p(not heads) = 1-h = j

    Ponder this (d+f)*(h+j) = d*h + d*j + f*h + f*j

    Four items to ponder. Which fit the given criteria?
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  3. #3
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    the or means both, but im not understanding what you want me to ponder on? or i should ask, what 4 items do you want me to ponder on?
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  4. #4
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    Quote Originally Posted by TKHunny View Post
    You must first decide what "or" means. Does it include both or is it exclusive?

    p(6) = d
    p(not six) = 1-d = f

    p(heads) = h
    p(not heads) = 1-h = j

    Ponder this (d+f)*(h+j) = d*h + d*j + f*h + f*j

    Four items to ponder. Which fit the given criteria?
    ---------------------------------------------------------------

    ok i have to be honest, but i have no idea what you just said to me....but nonetheless i figured out the answer to the problem...here is my solution.


    P(rolling a 6) = 1/6
    P(flipping a head) = 1/2
    P(rolling a 6 AND flipping a head) = 1/6 * 1/2 = 1/12

    P(rolling a 6 OR flipping a head) = 1/6 + 1/2 - 1/12 = 7/12


    or:

    1H, 1T, 2H, 2T, 3H, 3T, 4H, 4T, 5H, 5T, 6H, 6T

    Of those, the ones that are either a 6 or a head include

    1H, 2H, 3H, 4H, 5H, 6H, and 6T

    That's 7 out of 12
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  5. #5
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    Hello, slapmaxwell1!

    What is the probability of either a die roll of a 6
    or flipping a coin once and getting a Head?

    You are expected to know this formula:

    . . P(A \cup B) .= .P(A) + P(B) - P(A \cap B)


    We have: .P(6) = 1/6
    . . . . . . . .P(H) = 1/2
    . . . . . P(6 \cap H) = (1/6)(1/2) = 1/12


    Therefore:

    . . P(6 \cup H) .= .P(6) + P(H) - P(6 \cap H)

    . . . . . . . . .= .(1/6) + (1/2) - (1/12)

    . . . . . . . . .= . 7/12
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  6. #6
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    Quote Originally Posted by Soroban View Post
    Hello, slapmaxwell1!


    You are expected to know this formula:

    . . P(A \cup B) .= .P(A) + P(B) - P(A \cap B)


    We have: .P(6) = 1/6
    . . . . . . . .P(H) = 1/2
    . . . . . P(6 \cap H) = (1/6)(1/2) = 1/12


    Therefore:

    . . P(6 \cup H) .= .P(6) + P(H) - P(6 \cap H)

    . . . . . . . . .= .(1/6) + (1/2) - (1/12)

    . . . . . . . . .= . 7/12
    um thanks i know the formula now. :O)
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