Statistical Physics: Proving "if p(a)=p(b)=p then p(ab) ≤ p^2

[RUber]

how come this possible ?P(a/b) = P(a ∩ b) / P(b)

That's what it is.

 

[RUber]

You have the probability of a given b, so you are only looking at the subset of the possibilities where b has already happened. If the events are independent, the P(a ∩ b)=P(a)P(b), similarly P(a|b) = P(a). Test these against the formula, and you will see how it is true.

 

[rangatudugala]

 

[Ray Vickson]

[PLAIN]https://upload.wikimedia.org/math/9/9/b/99b11ba7fe4ec33be7a129cf182a32d2.png[/QUOTE]

Your formula $P(A \cap B) = P(A) \, P(B)$ is NOT a general, true formula; it is true if and only if $A$ and $B$ are "independent" events, such as getting 'heads' on toss 1 of a coin and getting 'tails' on toss 2, or successive particle emissions from a radioactive element. There are millions of real-world examples where it is false. (Or, maybe I mis-read the intent of your post, in which case you might have left out some crucial clarifying information.)

 

[HallsofIvy]

Do you not know that "P(a or b)= P(a)+ P(b)- P(a and b)"?

Also, way back in post #5, Ruber asked what "ab" meant and you never answered. Is it "a and b" or "a or b"?

 

[rangatudugala]

Do you not know that "P(a or b)= P(a)+ P(b)- P(a and b)"?

Also, way back in post #5, Ruber asked what "ab" meant and you never answered. Is it "a and b" or "a or b"?

I hv already answered. probably you didn't see

here rewritten the question
prove or disprove the following
p(a) = P(b) = p (let say some valve, you can use what ever symbol ) then p(a ∩ b) ≤ p^2