Question about conditional probability

[Alexsandro]

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).

 

[EnumaElish]

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).

 

[EnumaElish]

(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.