# Thread: proving one probability is greater than another

1. ## proving one probability is greater than another

The claim is : Let S be a well defined sample space where E and F are events such that
Pr(E) >1/2 implies Pr(F)<1/2

I have no background on proving probability theory. help please!!

2. Originally Posted by nikie1o2
The claim is : Let S be a well defined sample space where E and F are events such that
Pr(E) >1/2 implies Pr(F)<1/2
Of course that statement is false as written.
Consider tossing a single die.
Let E be the die shows at least three.
Let F be the die shows at most five.
$P(E)=\frac{4}{6}~\&~P(F)=\frac{5}{6}$.
Both are more than $\frac{1}{2}$.

Perhaps you misread the question or you left some bit of information out.

3. Yes, you are correct - a great detail...prove or DISPROVE the following claim. Thank you very much for the counter example

4. What would be the outcome if i had the same claim as above but in addition E and F are mutually exclusive ? i.e. prove or disprove- Let S be a well defined sample space where E and F are events such that E and F are mutually exclusive. Pr(E)>1/2 implies Pr(F)< 1/2

5. Originally Posted by nikie1o2
What would be the outcome if i had the same claim as above but in addition E and F are mutually exclusive ? i.e. prove or disprove- Let S be a well defined sample space where E and F are events such that E and F are mutually exclusive. Pr(E)>1/2 implies Pr(F)< 1/2
Indeed, if E and F are mutually exclusive then the statement is correct.
$1\ge P(E\cup F)=P(E)+P(F)>\frac{1}{2}+P(F)$.
So what about $P(F)?$

6. ## my counter example

Originally Posted by Plato
Of course that statement is false as written.
Consider tossing a single die.
Let E be the die shows at least three.
Let F be the die shows at most five.
$P(E)=\frac{4}{6}~\&~P(F)=\frac{5}{6}$.
Both are more than $\frac{1}{2}$.

Perhaps you misread the question or you left some bit of information out.
Pertaining to my first claim to prove or disprove: Let S be a well defined sample space where E and F are events such that
Pr(E) >1/2 implies Pr(F)<1/2

I don't know if this is right so maybe so can critique... As a counter example i said: Let S={1,2,3,4,5,6}. Let E={1,2,3,4} and Let F={1,2,3,4}. Then, the Pr(E)= 2/3 and the Pr(F)=2/3 which is >1/2 thus we have an implication with a true hypothesis and false conclusion which makes the implication false so it follows that the claim is also false. E.E.F