Q) If A and B are events in a probability space such that 0 < P(A) = P(B) = P (A intersect B) < 1, which of the following CANNOT be true?
I won't list all the options here, but the answer, according to ETS, is that "A and B are independent." But the other option in this questions says that "A is a proper subset of B," and this one is not a correct answer. So when can ever A be a proper subset of B, given the equality above?
Thanks.
GR 9367 #24 (probability)
Re: GR 9367 #24 (probability)
Given the standard normal distribution events denote the events (inf,0] and (inf,0] U {1} as A and B respectively. This gives that P(A)=P(B)=.5 and A is a proper subset of B. This example works b/c {1} has measure zero.
Re: GR 9367 #24 (probability)
If A,B are independent events, then P(A) times P(B) = P(A intersection B)
Since P(A)=P(B)=P(A intersection B), that would mean that to be independent P(A)=P(B)= P(A) times P(B). This could only occur if P(A)=P(B)= 0 or 1. However, the problem given said that 0<probabilities<1 .
Hence A,B cannot be independent.
BTW the post above mine is not correct, no offense. The probabilities do not have to necessarily be .5
Since P(A)=P(B)=P(A intersection B), that would mean that to be independent P(A)=P(B)= P(A) times P(B). This could only occur if P(A)=P(B)= 0 or 1. However, the problem given said that 0<probabilities<1 .
Hence A,B cannot be independent.
BTW the post above mine is not correct, no offense. The probabilities do not have to necessarily be .5

 Posts: 9
 Joined: Sat Jan 01, 2011 5:35 pm
Re: GR 9367 #24 (probability)
I realize this is quite the necro, but to anyone reading this, in the example given two posts above, the probabilities ARE .5, as it's half of a normal curve both times.
Don't listen to the last line of the above post.
Don't listen to the last line of the above post.
Re: GR 9367 #24 (probability)
Hey, I remember this post!
I think dasgut is right; I think that explanation shows that A can be a proper subset of B, and still have the same probability. I can understand soshdog's explanation for why A and B cannot be independent (in fact, I knew this before I asked the question), but that's not really what I asked for; I asked why one option is correct and the other one is incorrect.
The only explanation that I can see where A can be a proper subset of B is when there are uncountably many outcomes, as in dasgut's example.
I didn't bother to reply because I was actually taking the MGRE when soshdog posted his/her reply. But I would be happy to read more explanations (just for curiosity).
I think dasgut is right; I think that explanation shows that A can be a proper subset of B, and still have the same probability. I can understand soshdog's explanation for why A and B cannot be independent (in fact, I knew this before I asked the question), but that's not really what I asked for; I asked why one option is correct and the other one is incorrect.
The only explanation that I can see where A can be a proper subset of B is when there are uncountably many outcomes, as in dasgut's example.
I didn't bother to reply because I was actually taking the MGRE when soshdog posted his/her reply. But I would be happy to read more explanations (just for curiosity).

 Posts: 9
 Joined: Sat Jan 01, 2011 5:35 pm
Re: GR 9367 #24 (probability)
for any continous distribution, removing a single point from some interval [a,b] will impact the total probabilities not at all, as
$$\int_a^b f(x) dx=\int_a^c f(x)dx+ \int_c^b f(x) dx$$ for any $$c\in [a,b]$$
The left is "including" c, the right is not including it. Elementary calculus shows that this is the same.
You could actually pull out any set of measure zero, and this would still work out, but this is the simplest example, and one that doesn't involve the lesbauge integral.
$$\int_a^b f(x) dx=\int_a^c f(x)dx+ \int_c^b f(x) dx$$ for any $$c\in [a,b]$$
The left is "including" c, the right is not including it. Elementary calculus shows that this is the same.
You could actually pull out any set of measure zero, and this would still work out, but this is the simplest example, and one that doesn't involve the lesbauge integral.