Definition of dependence in probability

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Here is classical definition and example of dependent events.

"When two events are said to be dependent, the probability of one event occurring influences the likelihood of the other event. For example, if you were to draw a two cards from a deck of $52$ cards. If on your first draw you had an ace and you put that aside, the probability of drawing an ace on the second draw is greatly changed because you drew an ace the first time".

Let’s consider another scenario: Suppose I apply to a job. There are two interviews. The second interview ($B$) will take place only if I pass the first interview ($A$). So, we have probabilities of $P(A)$ and $P(B)$. Two events per se do not depend on each other because different people conduct the interviews. But $B$ will not take place if $A$ was a failure. So $P(B)\ne P(B\mid A)$. So, can it be said that events $A$ and $B$ are dependent?

Thanks!

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  1. "When two events are said to be dependent, the probability of one event occurring influences the likelihood of the other event."

  2. "Let’s consider another scenario: Suppose I apply to a job. There are two interviews. The second interview (B) will take place only if I pass the first interview (A). So, we have probabilities of P(A) and P(B). Two events per se do not depend on each other because different people conduct the interviews. But B will not take place if A was a failure."

Reasoning: Event B cannot occur without event A occurring before it.

Conclusion: A and B are dependent events.

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We say two events $A$ and $B$ are independent if knowing that one had occurred gave us no information about whether the other had occurred is

$$P(A|B) = P(A) $$

and

$$P(B|A) = P(B)$$

now we know

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

then

$$ P(A \cap B) = P(A)P(B)$$

clear if they do depend on each other this is not the case

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Perhaps the most concise definition of independent events is $P(A\,\text{and}\,B)=P(A)P(B)$. In your example, constants $p,\,q$ exist for which $P(A)=p,\,P(A\,\text{and}\,B)=P(B)=q$, with independence only if $p=1$.