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Thread: Question regarding inequalities and probability (Actex Problem Set 7 - Problem 3)

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    Question regarding inequalities and probability (Actex Problem Set 7 - Problem 3)

    In the solution for this problem, the probability for P[X + 10/x > 7] is simplified to P[(X - 5)(X - 2) > 0]. I can see how this is done by easily manipulating the inequality. However, the next step in the solution states:

    P[(X - 5)(X - 2) > 0] = P[X > 5] + P[X < 2]

    I don't understand why it's P[X < 2] and not P[X > 2]. Can someone clear this up for me? I'm sure it's an easy justification, but for some reason it's not obvious to me.

    Thanks!

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    Does X=3 or X=4 satisfy this equation? Once I got to P[(X-5)(X-2) > 0], I just plugged in all the numbers from 0 to 10 to see which would work. You'll see that it's numbers where X < 2 or X > 5.

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    Quote Originally Posted by bubbarobert8 View Post
    In the solution for this problem, the probability for P[X + 10/x > 7] is simplified to P[(X - 5)(X - 2) > 0]. I can see how this is done by easily manipulating the inequality. However, the next step in the solution states:

    P[(X - 5)(X - 2) > 0] = P[X > 5] + P[X < 2]

    I don't understand why it's P[X < 2] and not P[X > 2]. Can someone clear this up for me? I'm sure it's an easy justification, but for some reason it's not obvious to me.

    Thanks!
    Assume X=0.

    Then (X-5)(X-2) = (-5)(-2) = 10 > 0. This is an example where if X < 2, the (X-5)(X-2) > 0. It's a little counter intuitive, but Ross has the right idea essentially. It's because two negative real numbers multiplied by each other will equal a positive real number.

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    Thanks so much! A little plug and chug of the numbers makes it easier to understand. Guess I just thought there might be an obvious rule or something I was missing.

    I appreciate the help!!

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    You shouldn't need plug and chug in this case. As someone mentioned before a*b > 0 implies 2 possibilities

    1) a > 0 AND b > 0

    OR

    2) a < 0 AND b < 0

    So if you have (X-5)(X-2) > 0 then the 2 possibilities are

    1) X - 5 > 0 and X - 2 > 0 or in other words X > 5 AND X > 2 (that only happens when X > 5) i.e. the logical outcome of (X > 5) AND (X > 2) is X > 5

    OR

    2) X - 5 < 0 AND X - 2 < 0 or in other words X < 5 AND X < 2 once again this only happens when X < 2

    And of course in math AND means multiplication and OR means addition.

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