4
$\begingroup$

This is a basic question about probability theory.

My reasoning goes as follows:

  1. If $A$ and $B$ are independent events, the probability they both happen is their multiplication: $$\Pr(A \text{ and } B) = \Pr(A) \times \Pr(B)$$
  2. If their marginal probability is not impossible, also their product is non-zero: $$\Pr(A) > 0,\, \Pr(B) > 0 \implies \Pr(A \text{ and } B) > 0$$

  3. Hence, independent events cannot be disjoint

  4. Hence, only dependent events can be disjoint

  5. Hence, all disjoint events are dependent.

Can you help me point out the error in my argument?

$\endgroup$
7
  • 2
    $\begingroup$ Disjoint events are definitely not independent, you are right: knowing that one happens prevents the other from happening, so they are not independent. $\endgroup$
    – Randall
    Jan 24 '18 at 19:04
  • 3
    $\begingroup$ Technically, an event of probability 0 is independent of any other event, and it could also be disjoint from the other event. $\endgroup$ Jan 24 '18 at 19:07
  • 4
    $\begingroup$ You are right as long as A and B have positive probabilities $\endgroup$
    – Kun
    Jan 24 '18 at 19:20
  • $\begingroup$ So, if P(A and B) = 0 [ = disjoint], how can I calculate P(A|B), since they need to be dependent? $\endgroup$ Jan 24 '18 at 19:38
  • 1
    $\begingroup$ if $P(A\cap B)=0$ and $P(B)>0$ then $P(A|B)=0$. $\endgroup$
    – kludg
    Jan 24 '18 at 19:42
-1
$\begingroup$

Yes, as long as A and B are mutually exclusive and sum up to 1 or 100%, P(A) + P(B) = 1. Example : You are either dead or alive. The chance of being alive or dead always sums up tp 1 or 100%, no matter what condition you live in.

$\endgroup$

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.