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Few set and prob questions

  1. Mar 15, 2005 #1
    Well the first and last I'm having some troubles with, and 2-4 I think the logic I am using is correct but would like his verified since no answers were provided

    What is [itex]P(A \cup B)[/itex] if [itex] P(A) = 0.2, P(A \cap B) = 0.1, P(B) = 0.5?[/itex]

    Would that just be the prob. of being in A or B minus prob of being in both (prob of being in A + prob being in B - A int B). Would it depend on whether they are mutually exclusive or not? (how can we tell if thats all tahts given in the question).
    I am kind of half between (A + B) and half between (A + B - AintB). But since A int B was included in the question, would that imply that I should use A + B - A int B = 0.2 + 0.5 - 0.1 = 0.6

    ---
    What is E(X) if Mx(u) = [itex](1-u)^{-3}, u<1[/itex]

    To find E(X) find the first derivative:
    = -3(1-u)^(-4).-1
    = 3(1-u)^(-4)
    and then let u -> 0
    3(1)^(-4)
    =3

    Therefor E(X) = 3


    ---
    What is E([itex]X^{3}[/itex]) if fx(x) = 2x, 0<x<1

    E(X^(3)) = integral (0,1) of 2x.x^3 dx
    = int (0,1) 2x^4 dx
    = 2/5 x^5 .. (0,1)
    = 2/5

    Therefor E(X^3)) = 2/5

    ---
    What is c if
    g(x) = c|x|, x = -2, -1, 1, 2 is a probability function

    For it to be a prob. function, the sum of all the probabilities must equal 1
    2c + c + c + 2c = 1
    c = 1/6

    ---
    What is [itex] P(\overline{B})[/itex] if [itex] P(B|\overline{A}) = 0.5, P(\overline{A}) = 0.3[/itex] [itex] and P(B|A) = 0.8 ?[/itex]

    Well I'm a bit stuck on this question;
    I used some multiplicative laws to find
    [itex]P(A \cap B) = 0.56[/itex]
    and [itex] P(B \cap \overline{A}) = 0.15 [/itex]
    I'm not sure how to continue from here.

    Thanks
     
  2. jcsd
  3. Mar 15, 2005 #2

    xanthym

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    Science Advisor

    ITEM (A):
    P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
    ::: ⇒ P(A ∪ B) = (0.2) + (0.5) - (0.1)
    ::: ⇒ P(A ∪ B) = (0.6)

    ITEM (B):
    From problem statement:
    P(Ac) = (0.3)
    P(A) = 1 - P(Ac) = 1 - (0.3) = (0.7)

    P(B | A) = (0.8)
    P(Bc | A) = 1 - P(B | A) = 1 - (0.8) = (0.2)

    P(B | Ac) = 0.5
    P(Bc | Ac) = 1 - P(B | Ac) = 1 - (0.5) = (0.5)

    Thus, using above results:
    P(Bc) = P(Bc ∩ A) + P(Bc ∩ Ac) =
    = P(Bc | A)*P(A) + P(Bc | Ac)*P(Ac) =
    = (0.2)*(0.7) + (0.5)*(0.3)
    ::: ⇒ P(Bc) = (0.29)


    ~~
     
  4. Mar 15, 2005 #3

    HallsofIvy

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    Staff Emeritus
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    [itex]P(A\cup B)= P(A)+ P(B)- P(A \cap B)[/itex] so in this problem, yes, [itex]P(A\cup B)= .2+ .5- .1= 0.6[/itex].
    You know that A and B are not mutually exclusive because
    [itex]P(A\cap B)[/itex] is not 0!

    Assuming that Mx(u) is the moment generating function, then, yes, E(X) is the coefficient of u in the McLaurin expansion of Mx(u): 3 in this case.


    Yes, that's correct.

    Of course.

    If [itex]P(\overline A)= 0.3[/itex] then P(A)= 1- 0.3= 0.7
    [itex] P(B)= P(B|A)P(A)+ P(B|\overline A)P(\overline A)[/itex]
    = 0.8(0.7)+ 0.5(0.3)= 0.56+ 0.15= 0.71 so
    [itex]P(\overline B)= 1- 0.71= 0.29.
     
    Last edited: Mar 15, 2005
  5. Mar 16, 2005 #4
    Could someone please explain what [itex]A^c[/itex] means? I've never encountered this notation when working with sets? Is it the same as the cartesian product of a set: [itex]A^n[/itex]
     
  6. Mar 16, 2005 #5
    [tex]A^c[/tex] just means the complement of [tex]A[/tex].
     
  7. Mar 16, 2005 #6
    Ahh thank you all very much, I must have been unaware of the following results:
    [itex] P(B)= P(B \cap A) + P(B \cap \overline A)[/itex]
    [itex] P(B)= P(B|A)P(A)+ P(B|\overline A)P(\overline A)[/itex]
    and
    [itex] P(\overline B)= P(\overline B \cap A) + P(\overline B \cap \overline A)[/itex]
    [itex] P(\overline B)= P(\overline B|A)P(A)+ P(\overline B|\overline A)P(\overline A)[/itex]

    The prob. of B is the prob of B if A happens + the prob of B if a doesn't happen. Since either A either happens or it doesn't (duh).

    Once again thanks I understand it now.
     
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