1
$\begingroup$

Given that we know A and B are independent and they never occur at the same time, one of them must be impossible, no? $$ P(A\mid B)=\frac{P(A \cap B)}{P(B)}\\ \text{if A and B independent, B gives no additional information on the probability of A}\\ P(A\mid B) = P(A) = \frac{P(A \cap B)}{P(B)} \Rightarrow P(A) * P(B) = P(A \cap B) $$

If they then never occur at the same time, meaning $P(A \cap B) = 0$, either $A$ or $B$ must be impossible or not?

Edit: I know of the basics of conditional probability as used here. This is not part of any homework, just engagement with this on my own, to refresh my stats knowledge in combination with a learning about data analytics (Bayes' classifier).

Improve this question
$\endgroup$
4
  • $\begingroup$ I might be wrong but, isn't being independent and "not occurring at the same time" a contradictory statement? $\endgroup$
    – ffppnnr
    Commented May 8, 2019 at 12:54
  • $\begingroup$ Please add the [self-study] tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. $\endgroup$
    – gung - Reinstate Monica
    Commented May 8, 2019 at 13:40
  • $\begingroup$ @gung Added the self-study tag. I'm not really doing this for any class. And simply wanted to ask this question, as I saw it as odd. I'm not really doing this for class, but rather as part of a private endeavour on DataCamp. I have tried what I wrote and understand what I wrote, the basics of conditional probability. Would you mind removing the hold tag, so I can get some answers. $\endgroup$
    – thebilly
    Commented May 8, 2019 at 14:21
  • $\begingroup$ @ffppnnr You're almost, but not quite, correct. Consider the infinite sequences of iid Bernoulli$(p)$ variables. The set of sequences in which exactly one result is $1$ and the set of sequences in which exactly one result is $0$ obviously are disjoint ("not occurring at the same time") yet, by definition, are independent, because each event has probability zero and so does their intersection. I chose this example to emphasize that probability-zero events are not necessarily "impossible." $\endgroup$
    – whuber ♦
    Commented May 8, 2019 at 14:24

1 Answer 1

Reset to default
2
$\begingroup$

Correct. There is a fundamental contradiction here, and that lies in the fact that mutually exclusive (or incompatible) events cannot be independent, unless, as you suggest, we are dealing with events of measure $0$.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Could you explain what this 'measure of $0$' means ? $\endgroup$
    – thebilly
    Commented May 8, 2019 at 14:32
  • $\begingroup$ It means that the probability measure of an event $A$ is $0$ i.e. $P(A)=0$. In probability jargon, you will also hear the terms almost impossible or an event happening almost never. For further detail, I defer you to the excellent Wikipedia article. $\endgroup$
    – Easymode44
    Commented May 8, 2019 at 19:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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