Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Question about conditional probability

  1. Aug 25, 2005 #1
    Question about conditional probability. Can someone help me ?

    Repulsion. The event A is said to be repelled by the event B is P(A|B) < P(A), and to be attracted by B P(A|B) > P(A).

    (a) Show that if B attracts A, then A attracts B, and ~B repels A.

    (b) If A attracts B, and B attarcts C, does A attract C?

    (c) Explain this, throught ratio idea:

    P(A|B) > P(A) => P(B|A) > P(B).
     
  2. jcsd
  3. Aug 26, 2005 #2

    EnumaElish

    User Avatar
    Science Advisor
    Homework Helper

    Use definition of cond. prob., P(A|B) = P(A & B)/P(B).

    (a) B Attracts A ---> P(A & B)/P(B) > P(A) ---> P(A & B) > P(A)P(B) ---> P(A & B)/P(A) > P(B) ---> A attr. B.

    B Attracts A ---> P(A & B)/P(B) > P(A) ---> (P(A)-P(A & ~B))/(1-P(~B)) > P(A) ---> P(A)-P(A & ~B) > P(A) - P(A)P(~B) ---> -P(A & ~B) > - P(A)P(~B) ---> P(A & ~B) < P(A)P(~B) ---> P(A & ~B)/P(~B) < P(A) ---> ~B repels A.

    (c) See (a).
     
    Last edited: Aug 26, 2005
  4. Aug 26, 2005 #3

    EnumaElish

    User Avatar
    Science Advisor
    Homework Helper

    (b) Counterex.

    Let S = {1,2,2,3,4,5}
    A = {x is in S : x is even} = {2,2,4}
    B = {x is in S : x < 3} = {1,2,2}
    C = {x is in S : x is boldface} = {1,2,5}

    "P" is the uniform prob. measure over the elements of S.

    A and B attract each other.
    Proof: P(A & B) = P({2,2}) = 1/3 > 1/4 = P(A)P(B).

    B and C attract each other.
    Proof: P(B & C) = P({1,2}) = 1/3 > 1/4 = P(B)P(C).

    A and C do not attract each other.
    Proof: P(A & C) = P({2}) = 1/6 < 1/4 = P(A)P(C) ---> A and C repel.
     
    Last edited: Aug 26, 2005
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Question about conditional probability
Loading...