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Thread: Probability problem

  1. #1
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    Probability problem

    Could anyone suggest the solution for this problem?

    Consider Omega=[0,1] with probability defined on intervals by P([a,b])=b-a, 0<= a <= b <=1,


    1. Define a random variable by: X(omega)=min{omega,1-omega}.


    Find P({omega:X(omega)<= 4/5}), P({omega:X(omega)<=1/4}), P({omega:X(omega)<=0}), P({omega:X(omega)<=-1})
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  2. #2
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    Re: Probability problem

    The first probability is 1, as the maximum value for X is 0.5. The second probability will be 0.5. The third probability is 0. The last probability is 0.

    The first probability is $P([0,1]) = 1-0 = 1$.
    The second probability is $P\left( \left[ 0, \dfrac{1}{4} \right] \cup \left[ \dfrac{3}{4}, 1 \right] \right) = P\left( \left[ 0, \dfrac{1}{4} \right] \right) + P\left( \left[ \dfrac{3}{4}, 1 \right] \right) = \dfrac{1}{4} - 0 + 1 - \dfrac{3}{4} = \dfrac{1}{2}$
    The third probability is $P(\{0,1\}) = P([0,0]) + P([1,1]) = 0-0+1-1 = 0$
    The last probability is $P(\emptyset) = 0$.
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  3. #3
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    Re: Probability problem

    Quote Originally Posted by SlipEternal View Post
    The first probability is 1, as the maximum value for X is 0.5. The second probability will be 0.5. The third probability is 0. The last probability is 0.
    The first probability is $P([0,1]) = 1-0 = 1$.
    The second probability is $P\left( \left[ 0, \dfrac{1}{4} \right] \cup \left[ \dfrac{3}{4}, 1 \right] \right) = P\left( \left[ 0, \dfrac{1}{4} \right] \right) + P\left( \left[ \dfrac{3}{4}, 1 \right] \right) = \dfrac{1}{4} - 0 + 1 - \dfrac{3}{4} = \dfrac{1}{2}$
    The third probability is $P(\{0,1\}) = P([0,0]) + P([1,1]) = 0-0+1-1 = 0$
    The last probability is $P(\emptyset) = 0$.
    I am totally confused. What in the definition gives this: $P(\{0,1\}) = P([0,0]) + P([1,1])$ ?
    It seems to me in the O.P. $\omega$ is used in two different ways.

    Surely, $\mathcal{P}([0.2,0.7])=0.5$ But what does X(omega)=min{omega,1-omega} mean?
    "with probability defined on intervals by P([a,b])=b-a, 0<= a <= b <=1" WHAT?
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  4. #4
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    Re: Probability problem

    Quote Originally Posted by Plato View Post
    I am totally confused. What in the definition gives this: $P(\{0,1\}) = P([0,0]) + P([1,1])$ ?
    It seems to me in the O.P. $\omega$ is used in two different ways.

    Surely, $\mathcal{P}([0.2,0.7])=0.5$ But what does X(omega)=min{omega,1-omega} mean?
    "with probability defined on intervals by P([a,b])=b-a, 0<= a <= b <=1" WHAT?
    I took omega = $\omega$ while Omega = $\Omega$. So, $\omega \in [0,1] = \Omega$.

    Then, to evaluate a probability on a set containing only two points, I turned them into two intervals, as $[0,0] \cup [1,1] = \{0,1\}$
    Last edited by SlipEternal; Oct 10th 2017 at 11:32 AM.
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