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Conditional Probability & Independence

  1. Jan 30, 2010 #1
    1. The problem statement, all variables and given/known data

    One satellite is scheduled to be launched form Cape Canaveral in Florida, and another launching is scheduled for Vandenberg AFB in California. Let A denote the event that the Vandenberg launch goes off on schedule, and let B represent the event that the Cape Canaveral launch goes off on schedule. If A and B are independent events with P(A)>P(B) and P(A u B) = .626, P(A ∩ B) = .144, determine the values of P(A) and P(B).

    3. The attempt at a solution

    This is a practice problem. I have the answers of .45 & .32. I have no idea how to get them.

    I know that if P(A|B) = P(A), then the two probabilities are independent.

    I also know that P(A|B)P(B) = P(A ∩ B) .

    I don't know how to do this problem.
     
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  3. Jan 30, 2010 #2

    rock.freak667

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    Well then you know that P(A|B)P(B) = P(A ∩ B) simplifies to P(A∩B)=P(A)P(B).

    P(A u B) should be the same as P(A)+P(B)
     
  4. Jan 30, 2010 #3

    vela

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    That's not correct. You'd be double-counting when A and B occur simultaneously. It should be

    [itex]P(A \cup B)=P(A)+P(B)-P(A \cap B)[/itex].
     
  5. Jan 30, 2010 #4
    I actually got those two equations:

    [itex]
    P(A \cup B)=P(A)+P(B)-P(A \cap B)
    [/itex]
    and
    [itex]
    P(A\cap B)=P(A)P(B)
    [/itex]

    I end up with two equations with two unknowns, but I can't come up with any answers that work.
     
  6. Jan 30, 2010 #5

    vela

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    Solve for P(A) in the second equation and plug it into the first, then multiply through by P(B). You'll end up with a quadratic equation.
     
  7. Jan 30, 2010 #6
    Yup. And it comes out with no solutions. I've tried like 10 times.
     
  8. Jan 30, 2010 #7

    vela

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    You must be making an algebra error somewhere. I did the same thing and got the expected answer. Why don't you post your work so we can see where you might have gone astray?
     
  9. Jan 30, 2010 #8
    [itex]
    P(A \cup B)=P(A)+P(B)-P(A \cap B)
    [/itex]
    and
    [itex]
    P(A\cap B)=P(A)P(B)
    [/itex]

    [tex]
    P(A)=\frac{P(A\cap B)}{P(B)}
    [/tex]

    [tex]
    P(A \cup B)=\frac{P(A\cap B)}{P(B)}+P(B)-P(A \cap B)
    [/tex]

    [tex]
    P(A \cup B)=P(A\cap B)+P(B)^2-P(A \cap B)P(B)
    [/tex]

    [itex]
    .626=.144+P(B)^2-.144P(B)
    [/itex]

    [itex]P(B)^2-.144P(B)-.770=0[/itex]
     
  10. Jan 30, 2010 #9

    vela

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    You forgot to multiply the lefthand side by P(B).
     
  11. Jan 30, 2010 #10
    [tex]
    P(B)P(A \cup B)=P(A\cap B)+P(B)^2-P(A \cap B)P(B)
    [/tex]

    [tex].626P(B)=.144+P(B)^2-.144P(B)[/tex]
    [tex]0 = .144+P(B)^2-.770P(B)[/tex]

    Still no solutions.
     
  12. Jan 31, 2010 #11

    vela

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    Well, your equation is correct, and it has the aforementioned solutions. All I can suggest is you check your work again carefully.
     
  13. Jan 31, 2010 #12
    Oh. I'm a doofus. The coefficient of P(B)^2 is 1, not 0. Thanks for the help.
     
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