0
$\begingroup$

Suppose two events $A$ and $B$ are mutually exclusive then we say that $P$ ($A \cap B)=0$. In my knowledge event is the outcomes delivered when we preform a task. So in the above line, when two events are mutually exclusive, does that mean any outcome obtained in any one event is different from outcomes obtained in other event? Is that what we use to identify any two events (mutually exclusive, exhaustive etc ).

$\endgroup$
  • $\begingroup$ Not sure I understand your question. In the set up, one imagines that there is a trial and that, after the trial, we will be able to say whether a given event happened or not. To say the two events are mutually exclusive means that if you tell me that $A$ happened, then I know that $B$ did not and conversely (granted, for continuous distributions we may replace "did not happen" with "happened with probability $0$"). Is this what you are asking? $\endgroup$ – lulu Nov 10 '17 at 12:27
  • 3
    $\begingroup$ Yes, when you do a task (experiment) there are a set of possible outcomes S. Viewed this way, events are subsets of S. Mutually exclusive events (subsets) have no outcomes in common i.e an empty intersection in terms of subsets. Put in real terms, when one event happens the other cannot happen if they are mutually exclusive. $\endgroup$ – Paul Nov 10 '17 at 12:27
  • $\begingroup$ Just to be clear, given that the events are mutually exclusive, $A\cap B=\emptyset$ (the intersection is the empty set); but the probability is always a number, in this case $P(A\cap B) = 0$ (zero). It's not clear what you meant by "phie". $\endgroup$ – David K Nov 10 '17 at 12:32
  • $\begingroup$ an event is a subset of the sample space. See here. $\endgroup$ – Masacroso Nov 10 '17 at 12:41
2
$\begingroup$

Think of drawing one playing card from a shuffled deck.

The events drawing a 5 and drawing a 7 are "mutually exclusive" ... both cannot happen at the same time.

On the other hand, drawing a 5 and drawing a diamond are not mutually exclusive. They could both happen at the same time: if we draw the 5 of diamonds.

$\endgroup$
  • $\begingroup$ Basically an event happens if any one or more outcome of that event is associated with the sample space so the logic that drawing a 5 and getting diamond is not exclusive is bcos we can draw both of them at once which will mean they both are happening and hence are not exclusive ,right? $\endgroup$ – Hydrous Caperilla Nov 10 '17 at 14:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.