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    probability, independent or dependent




    Hello.

    I roll a dice twice. Let's say that A = odd number on first, B = even number on second, C = sum is odd.
    Now, I need to calculate P(A), P(B), P(C) and P(A cut B).

    Let's start, I think these three events are independent on each other.
    And thus these are ( I think ) relevant,

    P(A cut B) = P(A) P(B),
    P(A cut B) = P(A) P(C),
    P(B cut C) = P(B) P(C),
    P(A cut B cut C) = P(A) (B) P(C)

    cut means intersection, as you know.
    Because there are numbers 1-6,
    A = 1/6, B = 1/6, C = 1/6
    And therefore
    P(A) = P(A) P(B) = 1/6 * 1/6 = 1/36
    P(B) = P(A) P(C) = 1/6 * 1/6 = 1/36
    P(C) = P(B) P(C) = 1/6 * 1/6 = 1/36
    P(A cut B) = P(A cut B) = P(A cut C) = P(B cut C) = (1/6)^3 = 1/216

    Maybe this is not this easy? I don't know have I done this right?

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    Re: probability, independent or dependent

    Some Comments:

    A = odd number on first, B = even number on second, C = sum is odd.
    I think these three events are independent on each other.
    The assumption that A and B are independent is fair enough. C is pairwisely independent with A and B respectively but that is not immediate. Also note that they are not jointly/mutually independent.

    Because there are numbers 1-6,
    and the dice is fair so you have a discrete uniform distribution. But the events A, B, C are not asking you the probability of obtaining a specific number. So why 1/6?

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    Re: probability, independent or dependent

    I try again.

    Because there are odd or even numbers 1-3,
    A = 1/3, B = 1/3, C = 1/3

    And therefore

    P(A) = P(A) P(B) = 1/3 * 1/3 = 1/9 ( OR 1/3 + 1/3? )
    P(B) = P(A) P(C) = 1/3 * 1/3 = 1/9
    P(C) = P(B) P(C) = 1/3 * 1/3 = 1/9

    P(A cut B) = P(A cut B) = P(A cut C) = P(B cut C) = (1/3)^3

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    Re: probability, independent or dependent

    I think you need to start from the very basic concept before you blindly applying those properties...

    In a dice, how many faces are odd? how many are even? With the equally-likely assumption (the dice is fair), what is the probability of getting odd? and even?

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    Re: probability, independent or dependent

    Yes... of course... I was too hasty. Probability to get odd is 1 / 2 or even is 1/ 2.


    Because there are odd or even numbers
    A = 1/2, B = 1/2, C = 1/2

    And therefore

    P(A) = P(A) P(B) = 1/2 * 1/2 = 1/4 ( OR 1/2 + 1/2? )
    P(B) = P(A) P(C) = 1/2 * 1/2 = 1/4
    P(C) = P(B) P(C) = 1/2 * 1/2 = 1/4

    P(A cut B) = P(A cut B) = P(A cut C) = P(B cut C) = (1/2)^3

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    Re: probability, independent or dependent

    I don't know why you keep saying P(A) = P(A)P(B) which obviously does not hold unless P(B) = 1.

    The definition of independence is P(A \cap B) = P(A)P(B)

    And I think you can solve all with ease.

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    Re: probability, independent or dependent

    Okay, I try again.

    P(A) = 1/2
    P(B) = 1/2
    P(C) = 1/2

    P(A) cut P(B) = 1/2 * 1/2 = 1/4

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    Re: probability, independent or dependent


    I make a little adjustment here.

    P(A) = 1/2
    P(B) = 1/2
    P(C) = 1/2

    P(A cut B) = 1/2 * 1/2 = 1/4

    Are these now correct?

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