# Probability theory

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1. Aug 17, 2015

### rangatudugala

Prove or disprove the following statement:

If p(a)=p(b)=q then p(a∩b)≤q2

We know nothing know about event a , b.

3. The attempt at a solution

I tried this but don't know correct or not
Can some one help me
let a, b are independent event
0<q<1
then p(a∩b) = p(a) p(b) = q*q = q^2

2. Aug 17, 2015

### DEvens

3. Aug 17, 2015

### RUber

4. Aug 17, 2015

### rangatudugala

previous thread I used confusing notation (It was my 1st attempt )

5. Aug 17, 2015

### rangatudugala

I tried but no idea feel something missing
let a, b are independent event
0<q<1
then p(a∩b) = p(a) p(b) = q*q = q^2

6. Aug 17, 2015

### RUber

You won't prove that the statement is true or untrue by assuming that a and b are independent.
If a and b are independent, then the statement is true.
If a and b are not independent, you still don't know.
You either have to prove that
for all a and b such that p(a) = p(b) = q the statement is true
or
there exists at least one pair of a and b such that p(a) = p(b) = q AND the statement is false.
Let a = b and q = .5.
What do you find out?

7. Aug 17, 2015

### rangatudugala

Okay..
then p(a∩b) = p(a) p(b) =0.5*0.5 = 0.25

8. Aug 17, 2015

### rangatudugala

is there any other answer ??

9. Aug 18, 2015

### haruspex

No, that only works if they are independent. As RUber says, you need to think about two events that have some dependency. What's the most dependent you can make two events? (When two events have some dependency, knowing the outcome of one gives you information about the likely outcome of the other.)

10. Aug 18, 2015

### rangatudugala

p(a∩b) = p(a) + p(b) - p(ab)

Then how to get p(ab)??

11. Aug 18, 2015

### RUber

What is p(a ∩ a)?

12. Aug 18, 2015

### rangatudugala

p(a ∩ a) = 0.5
:?

13. Aug 18, 2015

### rangatudugala

isnt it?

14. Aug 18, 2015

### RUber

Right. Is that less than or equal to q^2?
Is p(a)=p(a) =q?

15. Aug 18, 2015

### rangatudugala

its greater than q^2

16. Aug 18, 2015

### rangatudugala

p(a)=p(a) =q yes

!!

17. Aug 18, 2015

### RUber

So then...?

18. Aug 18, 2015

### rangatudugala

1/ let a, b are independent event
0<q<1
then p(a∩b) = p(a) p(b) = q*q = q^2

2/ a, b not independent event
0<q<1

then p(a∩a) = q >= q^2

but why we going to calculate p(a∩a)??

question was p(a∩b) =< q^2 know

i confused again.. Am I missed step ?

19. Aug 18, 2015

### RUber

Okay, then let a and b be two separate events, that are highly correlated.
Let a = Is a student
Let b = Goes to school
Essentially, these are the same thing as saying a = b.

The reason we are doing this is because nothing in the statement you are trying to prove or disprove says we can't. If the statement is true, it must hold for all choices of a and b, including a = b. So if it doesn't hold for a = b, then it is disproved.
To disprove the statement, you don't have to show that it is always wrong, just that it could be wrong.
The fact is that the statement is true only when a and b are either independent or less likely to occur together than on their own.

So, like I have been saying. Use a = b and 0<q<1. Show that the example fits the assumptions for your statement (if ... ), then show that the outcome is false (then...) . This is how you disprove an if, then statement.

20. Aug 18, 2015

### rangatudugala

Big Thanks Sir...
Now I got the Point
we don't know anything about event a & b so we can do both prove & disprove using "if statements.

Thank you sir again.
Big sorry if I bothered to you so much.. :)