Proving p(ab) ≤ p² when p(a)=p(b)=p

[RUber]

Where are you going with this?
Assuming independence will change the problem.
We already covered that.
You don't know anything about P(a|b), so it can be anywhere from 0 to 1.

 

[rangatudugala]

sorry .. i dnt kw homework to prove this .. there may be a mistake some where or i dnt know nothing

Thanks for your help.. Ill try next prob

 

[RUber]

I have already told you a few times how to prove it.
If it were true that " if P(a) =P(b) =p then P(a ∩ b) ≤ p²" then the relation would have to hold for all values of p and all a and b such that P(a) =P(b) =p.
You have shown that it is true for independent a and b, and for mutually exclusive a and b. I will also add in that it is true for certain p =1.
It is certainly not true for any a and b such that 0<p<1 and p(a|b) > p.

 

[RUber]

independent
also P(a/b) = P(a)* P(a ∩ b) / P(b) ---> 3
P(b/a) = P(b)* P(a ∩ b) / P(a) ---> 4

Are you sure about this?
Your notation P(a/b) means probability of a given b, right?

This implies that P(a/b) = P(a ∩ b). Shouldn't it just be P(a/b) = P(a ∩ b) / P(b)?

 

[rangatudugala]

Are you sure about this?
Your notation P(a/b) means probability of a given b, right?

This implies that P(a/b) = P(a ∩ b). Shouldn't it just be P(a/b) = P(a ∩ b) / P(b)?

P(a/b) means probability of a given b, right? yes its true

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

 

[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