# Can two events in probability be both non-disjoint AND non-independent? Confused about interpretation of conditional probabilities?

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?)

• throw a die: the event "getting a six" and the event "getting an even number" are non-disjoint and non-independent – Masacroso Dec 27 '20 at 1:10
• 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. – bof Dec 27 '20 at 2:58

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 $$P(\text{first card is ace}) P(\text{second card is ace} \mid \text{first card is ace}) = \frac{4}{52} \cdot \frac{3}{51}$$.
• 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? – user865043 Dec 27 '20 at 3:25
• @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. – angryavian Dec 27 '20 at 4:00