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Math Help - Conditional probability

  1. #1
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    Conditional probability

    I have a sequence of states (either 1 or 2).

    For eg: Out of 100, if 30 of them are '1' and 70 of them are '2' then the P(1)=.3 and P(2)=.7

    Now I need to find the probability of occurance of 1 in the next state when the current state is already 1 P_(11). Also find the probability of occurance of 2 when the current state is 1 (P_12).

    Please help.
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  2. #2
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    Re: Conditional probability

    Quote Originally Posted by lemontree45 View Post
    I have a sequence of states (either 1 or 2).

    For eg: Out of 100, if 30 of them are '1' and 70 of them are '2' then the P(1)=.3 and P(2)=.7

    Now I need to find the probability of occurance of 1 in the next state when the current state is already 1 P_(11). Also find the probability of occurance of 2 when the current state is 1 (P_12).
    Please help.
    Is that example the actual probabilities you were given, or were you just trying to illustrate that the states 1 and 2 are exhaustive? Is it fair to assume that the next state is independent of the previous state? What is it about this problem specifically that you are having trouble with?
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  3. #3
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    Re: Conditional probability

    Hello, lemontree45!

    I have a sequence of states (either 1 or 2).

    For eg: Out of 100, if 30 of them are '1' and 70 of them are '2',
    then the P(1) = 0.3 and P(2) = 0.7

    Now I need to find the probability of occurance of 1 in the next state
    when the current state is already 1: P(11).
    Also find the probability of occurance of 2 when the current state is 1: P(12).

    I will ignore the reference to "100" and assume independent probabilities.
    . . That is, P(1) is always 0.3 . . . and P(2) is always 0.7.

    Then we have:

    . . \begin{Bmatrix}P(11) \:=\:(0.3)(0.3) \:=\:0.09 && P(12) \:=\:(0.3)(0.7) \:=\:0.21 \\ \\[-3mm] P(21) \:=\:(0.7)(0.3) \:=\:0.21 && P(22) \:=\:(0.7)(0.7) \:=\:0.49 \end{Bmatrix}

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