Describing the relationship between two sets A and B (probability)

[Byeongok]

Hi I am new here!
hopefully someone is kind enough to reply fast and help.

so the question I am stuck is:

Describe the relationship between two sets A and B ( A and B are non-empty) if:
a. Pr(A|B)=Pr(A)
b. Pr(A/B)=0
c. Pr(A/B)=Pr(A)/Pr(B)

(Sorry guys can't get the fraction signs working! so / means fraction or divide)

Usually i'd add some working that i did, but i just can't understand the question it self and the notations used.

 

[caffeinemachine]

Hi I am new here!
hopefully someone is kind enough to reply fast and help.

so the question I am stuck is:

Describe the relationship between two sets A and B ( A and B are non-empty) if:
a. Pr(A|B)=Pr(A)
b. Pr(A/B)=0
c. Pr(A/B)=Pr(A)/Pr(B)

(Sorry guys can't get the fraction signs working! so / means fraction or divide)

Usually i'd add some working that i did, but i just can't understand the question it self and the notations used.

Hello Year10Student. Welcome to the forum.

Can you please write out the definition of $Pr(A/B)$??

I know that $Pr(A|B)$ si defined as $Pr(A\cap B)/Pr(B)$. Substituting this, you get in part (a) that $Pr(A\cap B)=Pr(A)Pr(B)$. Thus, by definition, we conclude that $A$ and $B$ are independent events.

If you could tell us the definition of $Pr(A/B)$ then we can help you with the remaining two parts.

 

[Byeongok]

I am not sure but i think it means the same as Pr (A|B)

Thanks!

 

[caffeinemachine]

I am not sure but i think it means the same as Pr (A|B)

Thanks!

If so, does the solution to part (a) give you any ideas?

 

[Byeongok]

If so, does the solution to part (a) give you any ideas?

I'm not sure if its correct but ill have ago again!

EDIT:

a.

Pr (A and B) / Pr (B) = Pr (A)
Events A and B are independent.
Probability of A occurring doesn't affect the probability of B happening.


b.

P (A and B) / P (B) = 0
A and B are mutually exclusive.
The probability of A and B together equals 0. Meaning they can’t happen together.

c.

P (A and B) / P (B) = P (A) / P (B)
A is a subset of B.

 

[caffeinemachine]

I'm not sure if its correct but ill have ago again!

EDIT:

a.

Pr (A and B) / Pr (B) = Pr (A)
Events A and B are independent.
Probability of A occurring doesn't affect the probability of B happening.


b.

P (A and B) / P (B) = 0
A and B are mutually exclusive.
The probability of A and B together equals 0. Meaning they can’t happen together.

c.

P (A and B) / P (B) = P (A) / P (B)
A is a subset of B.

Yes that's correct!

 

[Byeongok]

yes that's correct!

thanks caffeinemachine!

 

[johng1]

Hi,
For part c), you really can't conclude A is a subset of B. All you can say is that $P(A\setminus B)=0$. That is the probability that A happens and B doesn't is 0. Example: Let the probability space be a square of unit area with the probability of a sub region of the square being the area of the region. Then take a triangle of area one half as a region, including the boundary edges. Imagine throwing a dart at the square. The probability of hitting the triangle is 1/2. Now take B to be the interior of the triangle. The probability of hitting B is also 1/2; i.e. $P(A\cap B)=P(A)$. Clearly A is not a subset of B. This just means the probability of hitting anyone of the 3 edges (each of area 0) is 0.

 

[caffeinemachine]

Hi,
For part c), you really can't conclude A is a subset of B. All you can say is that $P(A\setminus B)=0$. That is the probability that A happens and B doesn't is 0. Example: Let the probability space be a square of unit area with the probability of a sub region of the square being the area of the region. Then take a triangle of area one half as a region, including the boundary edges. Imagine throwing a dart at the square. The probability of hitting the triangle is 1/2. Now take B to be the interior of the triangle. The probability of hitting B is also 1/2; i.e. $P(A\cap B)=P(A)$. Clearly A is not a subset of B. This just means the probability of hitting anyone of the 3 edges (each of area 0) is 0.

Good catch. This is correct. I am sorry I missed that.