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Thread: Probability Question

  1. #1
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    Probability Question

    Question :

    Let $\displaystyle X_1$ and $\displaystyle X_2$ be the joint probability distribution

    $\displaystyle . \qquad \qquad \qquad \qquad X_1=0$ $\displaystyle . \qquad \qquad \qquad X_1=1$ $\displaystyle . \qquad \qquad \qquad X_1=2$
    $\displaystyle X_2 = 0$ $\displaystyle . \qquad \qquad \qquad 0.1$ $\displaystyle . \qquad \qquad \qquad \qquad 0.4$ $\displaystyle . \qquad \qquad \qquad \qquad 0.1$
    $\displaystyle X_2 = 1 $ $\displaystyle . \qquad \qquad \qquad 0.2$ $\displaystyle . \qquad \qquad \qquad \qquad 0.2$ $\displaystyle . \qquad \qquad \qquad \qquad 0.0$

    i) Find P(X_1 + X_2)
    ii) Are X_1 and X_2 independent
    iii) P(X_1 = 0 | X_2 = 1)

    I need to know how to get the value of $\displaystyle P(X_1 + X_2)$

    Here is my work
    $\displaystyle
    \begin{array}{|c|c|c|} P(X_1 = 0) \ = \ 0.3 \ \ & P(X_1 = 0 |X_2 = 0) \ = \ 0.2 \ \ & P(X_1 = 0 | X_2 = 1) \ = \ 0.5 \\
    P(X_1 = 1) \ = \ 0.6 \ \ & P(X_1 = 1 |X_2 = 0) \ = \ 0.7 \ \ & P(X_1 = 1 | X_2 = 1) \ = \ 0.5 \\
    P(X_1 = 2) \ = \ 0.1 \ \ & P(X_1 = 2 |X_2 = 0) \ = \ 0.2 \ \ & P(X_1 = 2 | X_2 = 1) \ = \ 0.0 \\
    \end{array}
    $

    And also want to know the if my answers of ii) and iii) are correct or no ?

    i) ???

    ii) Answer
    Since $\displaystyle P(X_1|X_2) = P(X_1) \ \therefore \ P(X_1) \ and \ P(X_2)$ are not independent
    iii) Answer
    $\displaystyle
    P(X_1 = 0| X_2 = 1) \ = \ 0.5$

    Please provide me with the answer for i) ?????????
    Last edited by mr fantastic; Dec 10th 2009 at 02:12 PM. Reason: Corrected a typo (in red)
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  2. #2
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    Im confused. Are they asking you to find the $\displaystyle P(X_1,X_2)$, or literally the value of $\displaystyle X_1$ added to $\displaystyle X_2$, which is a random variable from 0 to 3?

    As for independence, they are most surely not. Take the probability of (0,0):

    $\displaystyle P(X_1=0,X_2=0) \neq P(X_1=0)P(X_2=0)$

    As the probability on the left is 0.1, however the probability on the right is 0.18.

    Your conditional probability is correct.
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  3. #3
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    Quote Originally Posted by zorro View Post
    Question :

    Let $\displaystyle X_1$ and $\displaystyle X_2$ be the joint probability distribution

    $\displaystyle . \qquad \qquad \qquad \qquad X_1=0$ $\displaystyle . \qquad \qquad \qquad X_1=1$ $\displaystyle . \qquad \qquad \qquad X_1=2$
    $\displaystyle X_2 = 0$ $\displaystyle . \qquad \qquad \qquad 0.1$ $\displaystyle . \qquad \qquad \qquad \qquad 0.4$ $\displaystyle . \qquad \qquad \qquad \qquad 0.1$
    $\displaystyle X_2 = 1 $ $\displaystyle . \qquad \qquad \qquad 0.2$ $\displaystyle . \qquad \qquad \qquad \qquad 0.2$ $\displaystyle . \qquad \qquad \qquad \qquad 0.0$

    i) Find P(X_1 + X_2)
    ii) Are X_1 and X_2 independent
    iii) P(X_1 = 0 | X_2 = 1)

    I need to know how to get the value of $\displaystyle P(X_1 + X_2)$

    Here is my work
    $\displaystyle
    \begin{array}{|c|c|c|} P(X_1 = 0) \ = \ 0.3 \ \ & P(X_1 = 0 |X_2 = 0) \ = \ 0.2 \ \ & P(X_1 = 0 | X_2 = 1) \ = \ 0.5 \\
    P(X_1 = 1) \ = \ 0.6 \ \ & P(X_1 = 1 |X_2 = 0) \ = \ 0.7 \ \ & P(X_1 = 1 | X_2 = 1) \ = \ 0.5 \\
    P(X_1 = 2) \ = \ 0.1 \ \ & P(X_1 = 2 |X_2 = 0) \ = \ 0.2 \ \ & P(X_1 = 2 | X_2 = 1) \ = \ 0.0 \\
    \end{array}
    $

    And also want to know the if my answers of ii) and iii) are correct or no ?

    i) ???

    ii) Answer
    Since $\displaystyle P(X_1|X_2) = P(X_1) \ \therefore \ P(X_1) \ and \ P(X_2)$ are not independent
    iii) Answer
    $\displaystyle
    P(X_1 = 0| X_2 = 1) \ = \ 0.5$

    Please provide me with the answer for i) ?????????
    Your answer for (iii) is correct: 0.2/(0.2 + 0.2 + 0.0) = 1/2.

    For (i), I would let $\displaystyle Y = X_1 + X_2$. The possible values for Y are Y = 0, 1, 2, 3. Your job is to calculate Pr(Y = 0), Pr(Y = 1) etc. eg. $\displaystyle \Pr(Y = 1) = \Pr(X_1 = 0 \cap X_2 = 1) + \Pr(X_1 = 1 \cap X_2 = 0) = 0.2 + 0.4 = 0.6$ etc.

    For (ii), it's sufficient to show something like $\displaystyle \Pr(X_1 = 0 \cap X_2 = 0) \neq \Pr(X_1 = 0) \cdot \Pr(X_2 = 0)$.
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  4. #4
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    I wasnt able to understand

    Quote Originally Posted by mr fantastic View Post
    Your answer for (iii) is correct: 0.2/(0.2 + 0.2 + 0.0) = 1/2.

    For (i), I would let $\displaystyle Y = X_1 + X_2$. The possible values for Y are Y = 0, 1, 2, 3. Your job is to calculate Pr(Y = 0), Pr(Y = 1) etc. eg. $\displaystyle \Pr(Y = 1) = \Pr(X_1 = 0 \cap X_2 = 1) + \Pr(X_1 = 1 \cap X_2 = 0) = 0.2 + 0.4 = 0.6$ etc.

    For (ii), it's sufficient to show something like $\displaystyle \Pr(X_1 = 0 \cap X_2 = 0) \neq \Pr(X_1 = 0) \cdot \Pr(X_2 = 0)$.

    Mr fantastic i wasnt able to understand the answer 1)
    could u please tell me why is Y used at all
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  5. #5
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    Quote Originally Posted by zorro View Post
    Mr fantastic i wasnt able to understand the answer 1)
    could u please tell me why is Y used at all
    For convenience. I have defined Y to be equal to X1 + X2. If you don't like it, just continue using X1 + X2.
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    $\displaystyle Y$ is the random variable whose values are the values of the random variable $\displaystyle X_1$ and $\displaystyle X_2$ added together. Therefore $\displaystyle Y$ is just the possible combinations of $\displaystyle X_1$ added to $\displaystyle X_2$, and has values that range from 0 to 3.
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  7. #7
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    Question

    Quote Originally Posted by mr fantastic View Post
    For convenience. I have defined Y to be equal to X1 + X2. If you don't like it, just continue using X1 + X2.

    May i know what is $\displaystyle P(X_1 + X_2)$ is it $\displaystyle P(X_1) + P(X_2) $ or some thing else

    In ur prev post i wasnt able to understand this bit
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    Despite your disbelief in the powers of others: As Mr. Fantastic (and myself) said, Y is simply the possible values of X1 (which are 0, 1 or 2), added to the possible values of X2 (which are 0, 1). Therefore Y is a random variable, made from adding two random variables together:

    0 + 0 = 0, 0 + 1 = 1, 1 + 0 = 1, 1 + 1= 2, 2 + 0= 2, 2 + 1 = 3;

    are all the possible combinations of the two random variables X1 and X2.

    It really is that literal.
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  9. #9
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    Quote Originally Posted by zorro View Post
    May i know what is $\displaystyle P(X_1 + X_2)$ is it $\displaystyle P(X_1) + P(X_2) $ or some thing else

    In ur prev post i wasnt able to understand this bit
    X1 + X2 is a random variable (which I called Y). $\displaystyle P(X_1 + X_2)$ is the probability of that random variable being equal to the values that it can equal. eg. Pr(X1 + X2 = 0), Pr(X1 + X2 = 1), ..... Pr(X1 + X2 = 3).
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  10. #10
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    Is this correct?

    Quote Originally Posted by mr fantastic View Post
    X1 + X2 is a random variable (which I called Y). $\displaystyle P(X_1 + X_2)$ is the probability of that random variable being equal to the values that it can equal. eg. Pr(X1 + X2 = 0), Pr(X1 + X2 = 1), ..... Pr(X1 + X2 = 3).


    $\displaystyle P(X_1 + X_2 = 0) = 0.06$
    $\displaystyle P(X_1 + X_2 = 1) = 0.6$
    $\displaystyle P(X_1 + X_2 = 2) = 0.32$
    $\displaystyle P(X_1 + X_2 = 3) = 0.0$
    $\displaystyle \therefore$ $\displaystyle P(X_1 + X_2) = 0.98$


    Is this right ????
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    The probability that Y takes on some value between 0 and 3 is 1. You are being asked to construct a probability distribution for P(X1+X2).

    Also, you should know that 0.98 isn't right simply by the fact that the probability of P(S) (where S is your space) is equal to 1.
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  12. #12
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    Quote Originally Posted by zorro View Post
    $\displaystyle P(X_1 + X_2 = 0) = 0.06$
    $\displaystyle P(X_1 + X_2 = 1) = 0.6$
    $\displaystyle P(X_1 + X_2 = 2) = 0.32$
    $\displaystyle P(X_1 + X_2 = 3) = 0.0$
    $\displaystyle \therefore$ $\displaystyle P(X_1 + X_2) = 0.98$


    Is this right ????
    $\displaystyle P(X_1 + X_2 = 0) = P(X_1 = 0 \cap X_2 = 0) = 0.1$ straight from the given table.

    $\displaystyle P(X_1 + X_2 = 2) = P(X_1 = 1 \cap X_2 = 1) + P(X_1 = 2 \cap X_2 = 0) + = .... $ (get all values from the given table).

    etc.
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