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I am currently reviewing conditional probability in my textbook and have stumbled across this:

The formula for conditional probability is this: Where $A$ and $B$ are two events where $P(B) > 0$ (naturally)

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

so naturally: $P(A\cap B) = P(B)\cdot P(A\mid B)$

and if $A$ and $B$ are independent, $P(A \mid B) = P(A)$

so $P(A\cap B) = P(B) \cdot P(A)$ which matches the definition of independence.

What I wonder is this. Say that $P(B)$ and $P(A \mid B)$ are non-zero probabilities and $P(A \mid B) \neq P(A)$ so $P(A\cap B) \neq 0$ and $A$ and $B$ are non-disjoint AND not independent events. Is this legal? Have I broken a law of probability? The textbook doesn't specify. If it's legal can someone give me an example in real life of two events that are non-disjoint and not independent?

Also can someone answer my second question?

Does the RHS of $P(A\cap B) = P(B)\cdot P(A\mid B)$ mean that event $B$ occurs first and THEN event $A$ occurs in which event $B$ occurring first affected the probability of event $A$ also occurring?

OR

Does the RHS mean that event $B$ and $A$ occur simultaneously? (If it's this second interpretation, can someone explain how events $A$ and $B$ occurring simultaneously affects event $A$'s probability of occurring?)

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    $\begingroup$ throw a die: the event "getting a six" and the event "getting an even number" are non-disjoint and non-independent $\endgroup$ – Masacroso Dec 27 '20 at 1:10
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    $\begingroup$ Throw a die. The events "get a one" and "get a two" are disjoint and not independent. The complementary evens, "not get a one" and "not get a two", are not disjoint and not independent. $\endgroup$ – bof Dec 27 '20 at 2:58
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Any example where $A \subsetneq B$ (with $P(A) > 0$ and $P(B) < 1$) will satisfy your conditions:

  • $A$ and $B$ are clearly not disjoint
  • $P(A \cap B) = P(A) \ne P(A) P(B)$

$P(A \mid B)$ should be interpreted as the probability of $A$ occurring if you already know that $B$ occurred. So you can interpret $P(B) \cdot P(A \mid B)$ as first accounting for the probability that $B$ occurred, and then accounting for the probability that $A$ occurred given that you already know $B$ occurred. This is common in sequential computations, i.e. the probability that two draws without replacement from a deck are both aces is $P(\text{first card is ace}) P(\text{second card is ace} \mid \text{first card is ace}) = \frac{4}{52} \cdot \frac{3}{51}$.

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  • $\begingroup$ Thank You. But I still have a question about the second part. Would $P(A\cap B) = P(B) \cdot P(A|B)$ be logically equivalent to $P(A) \cdot P(B|A) $? Despite the order of events occurring in the sequence being different? An example being: Is the probability of someone getting hit by a train and then struck by lighting logically equivalent to the probability of someone getting struck by lighting and then hit by a train? Does order matter? $\endgroup$ – user865043 Dec 27 '20 at 3:25
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    $\begingroup$ @BillBillwater They're mathematically equivalent, in that they both equal $P(A \cap B)$. I'm not sure what "logically equivalent" means. But you shouldn't think of the conditioning as necessarily sequential (see the examples given by Masacroso and bof). The train and lightning question is a little too vague for me to answer mathematically. $\endgroup$ – angryavian Dec 27 '20 at 4:00
  • $\begingroup$ Oh sorry, when I say logically equivalent, I just mean equivalent, but I see from your response that they're equal and that whether you think of it as sequential or not all just depends on the context of the situation. Thank You! $\endgroup$ – user865043 Dec 27 '20 at 4:07
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It not only is not illegal but it is the typical situation. For example: The event that a man is more than six feet tall and the event that he weighs more than 200 pounds are not independent and not disjoint.

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