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First we have that the probability of $A$ occurring given that $B$ occurs is: $$P(A|B) = \frac{P(A \cap B)}{P(B)}$$ but if the events are independent then that would simply equal $P(A)$ so what's the use of this formula? where do we need it? And I can't seem to make sense of the whole formula if the events are dependent so I can't really understand it. (professor didn't mention anything about what type of events this is used with)

RobPratt
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Sergio
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  • conditional probability makes sense even when the events are dependent. "What's the probability that $X$ has a given disease given that they have some symptoms of the disease?", that sort of thing. – lulu Sep 14 '20 at 15:02
  • I meant in terms of formula and how it came up not in terms of usage for dependent events @lulu – Sergio Sep 14 '20 at 15:11
  • I don't understand. The formula is the definition of conditional probability. What are you asking? – lulu Sep 14 '20 at 15:12
  • How they came up with it and such – Sergio Sep 14 '20 at 15:15
  • It's common sense. if I wanted the answer to the question I asked in my first comment, I would (in theory) count the number of people who had the given symptoms, then count the number of those who also had the given disease, and divide the latter by the former. That's the formula, more or less. To change the count into probabilities you divide both numerator and denominator by the total population, but that factor cancels out, – lulu Sep 14 '20 at 15:17

2 Answers2

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You are correct, if $A$ and $B$ are independent, then $P(A\cap B) = P(A)P(B)$ and the entire expression is not very enlightening.

But when $A$ and $B$ are dependent, this result becomes very meaningful. For example, consider this problem: With the probability of $1/3$, exactly one of eight identical-looking envelopes contains a bill (conditional probability question) or the very famous Monty Hall puzzle.

gt6989b
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You can use the formula whenever you want. If the events are independent you will have

$$\mathbb{P}[A|B]=\frac{\mathbb{P}[A]\times \mathbb{P}[B]}{\mathbb{P}[B]}=\mathbb{P}[A]$$

So you used the same formula...but with a simplified result

tommik
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