Intro statistics question: probability of intersection

[Eats Dirt]

Homework Statement

If event A equals event B, then the probability of their intersection is 1. True or False?

Apparently the correct answer is False.

The Attempt at a Solution

If A=B then they should overlap entirely and their intersection should be 1? The only way I see this working is if A is a subset of B and therefore they do not overlap completely, but when the question states "equals" I would think this means they are the same.

 

[Ray Vickson]

Homework Statement

If event A equals event B, then the probability of their intersection is 1. True or False?

Apparently the correct answer is False.

The Attempt at a Solution

If A=B then they should overlap entirely and their intersection should be 1? The only way I see this working is if A is a subset of B and therefore they do not overlap completely, but when the question states "equals" I would think this means they are the same.

Suppose that in one toss of a coin we have A ={get heads} and B = {do not get tails}. Do you agree that A=B? What is their intersection? Why do you think that their intersection is 100% certain?

 

[Eats Dirt]

Suppose that in one toss of a coin we have A ={get heads} and B = {do not get tails}. Do you agree that A=B? What is their intersection? Why do you think that their intersection is 100% certain?

Yes, I agree that A = B in this sense because there are two options, heads or tails - getting a head and not getting a tail is the same thing. By definition the intersection would be those elements that are in common between the events, in this case getting heads. It is 100% certain in terms of probability because the elements of each event (getting heads or not getting tails) are the same and thus their P(intersection) = 1. So isn't the statement true?

 

[Ray Vickson]

Yes, I agree that A = B in this sense because there are two options, heads or tails - getting a head and not getting a tail is the same thing. By definition the intersection would be those elements that are in common between the events, in this case getting heads. It is 100% certain in terms of probability because the elements of each event (getting heads or not getting tails) are the same and thus their P(intersection) = 1. So isn't the statement true?

So, I can be 100% sure to get a "head" in a coin-toss just by describing it in two ways? If I say "get heads", that has probability 1/2, and if I say "do not get tails" that has probability 1/2 also, but if I say it in two different ways it suddenly has probability 1?

 

[Eats Dirt]

So, I can be 100% sure to get a "head" in a coin-toss just by describing it in two ways? If I say "get heads", that has probability 1/2, and if I say "do not get tails" that has probability 1/2 also, but if I say it in two different ways it suddenly has probability 1?

But it is asking about the probability of their "Intersection" so shouldn't the intersection between the two overlap completely and be 1?

 

[Ray Vickson]

But it is asking about the probability of their "Intersection" so shouldn't the intersection between the two overlap completely and be 1?

That is why I asked you to tell me what is the intersection of the events A={get heads} and B = {do not get tails}. The intersection $A \cap B$ is some subset of the sample space $S = \{ H,T \}.$ What IS that subset? Do not tell me in words; actually display the subset.

 

[Eats Dirt]

That is why I asked you to tell me what is the intersection of the events A={get heads} and B = {do not get tails}. The intersection $A \cap B$ is some subset of the sample space $S = \{ H,T \}.$ What IS that subset? Do not tell me in words; actually display the subset.

Their intersection would be a subset of S say, $Sub = \{ H \}.$

 

[Eats Dirt]

Their intersection would be a subset of S say, $Sub = \{ H \}.$

ok i got it thank you

 

[Merlin3189]

For a pack of cards;

say, event A is getting a club, event B is getting a ten, What is the probability of their intersection?
In this case A≠B and P(A) ≠ P(B) ≠ P(A∩B)≠P(A) , P(A)=1/4 ,P(B) = 1/13 and P(A∩B) = 1/52

Now say event A is getting a red card and event B is getting a heart or a diamond. Then P(A) = P(B) = P(A∩B) = 1/2, since A, B and A∩B are all the same

May be you're thinking about the logical expression ( P(A)=P(B) ) which would be true if A=B, since P(A) would = P(B).
True and false are sometimes represented as 1 and 0.

 

[Ray Vickson]

ok i got it thank you

Right.

Of course, if $A = B$ then $A \cap B = A$, so $P(A \cap B) = P(A)$.

You may have been accidentally thinking of conditional probabilities, because for them it is true that
$$ A = B \; \; \Rightarrow \;\; P(A | B) = 1.$$