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Firstly I wanted to ask for this question how can probability of a and b add up to more then one. If this happens then are they saying the two events do not belong to the same sample space because that's the only way that their probabilities can add up to more than 1 I think.

Also, if we have two different events from two different experiments is it possible for us to compare them e.g to see if they are independent. E.g let one experiment be rolling a die once and another experiment be tossing a coin. Then if we choose an event from experiment 1 and an event from experiment two is it possible for us to compare the independence of those two events? (Although they would be independent anyways.)

Sorry if this seems like a stupid question.

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[iv] is necessarily false. – cjm Apr 30 '14 at 22:14
Regarding your first paragraph, For a concrete example: Roll a die. Let $A$ be the event "number is less than 5" and $B$ be "number is greater than 3". – Hugh Apr 30 '14 at 23:36

Consider a dice and the following events:

  1. $A$: the dice returns an even number $\left(P(A) = \frac{3}{6}\right)$
  2. $B$: the dice returns a number different from $6$ $\left(P(B) = \frac{5}{6}\right)$

Then $P(A) + P(B) = \frac{8}{6} > 1$.

Hope this can help you to understand that there are cases for which $P(A) + P(B) > 1$. This is because $A$ and $B$ overlaps ($A$ is satisfied for $2, 4$ and $6$, while $B$ is satisfied for $1, 2, 3, 4$ and $5$). Notice that overlapping does not always means that $P(A) + P(B) > 1$ (take $B$ as "the dice returns $1$ or $2$").

Anyway, if $P(A) + P(B) > 1$, you are sure that $A$ and $B$ overlap.

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$$P(A)+P(B)-P(A\cap B)=P(A\cup B)$$

So if events $A$ and $B$ overlap, it is possible for $P(A)+P(B) > 1$.

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$P(A|B)=P(B|A)$ means

$$\frac{P(A\cap B)}{P(B)}=\frac{P(A\cap B)}{P(A)}$$

So $P(A)=P(B)$, unless $P(A \cap B)=0$. But this can't happen, otherwise the sum $P(A)+P(B)$ would be $\leq 1$.

Thus, from $P(A)+P(B)>1$ you know that A and B overlap (so (iii) and (iv) are false), and from the other equality you know that $P(A)=P(B)$.

Then, if A and B are independant, you have $P(A \cap B)=P(A)P(B)$, so $$P(A \cup B) = P(A) + P(B) - P(A\cap B) = 2P(A) - P(A)^2=P(A)(2-P(A))$$

The maximum of $x(2-x)$ is $1$, found at $x=1$. So if A and B are independant, they are not exhaustive unless $P(A)=P(B)=1$.

And indeed, if $P(A)=P(B)=1$, your constraints are satisfied and (i) and (ii) are then true, so they may happen. But you could also have $A=B$ and $\frac12<P(A)=P(B)<1$, then $P(A \cap B)=P(A) \ne P(A)^2=P(A)P(B)$, so A and B are not independant, and they are not exhausive either because $P(A \cup B)= P(A) + P(B) - P(A \cap B)=P(A)<1$.

Hence (v) is true.

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I think if you let $A=B$ with $1/2<P(A)<1$, then you see that [v] is true.

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