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I am watching this video on YouTube to understand dependent and independent events and conditional probability. The example author uses is of a pair of dice.

There are 2 dice, Red and Green, let's say:

  • A is an event where Red die $\ge 4$
  • B is an event where number on Green die is larger than the number in event A.

then are A and B independent ?

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

$$P(B) = 15/36, P(A \cap B) = 3/36 \implies P(A|B) = 1/5$$

Now And and B independent only if $$P(A|B) = P(A)$$

and probability of Red die being greater than 4 is just $1/2$

$$ P(A | B) = P(A) \implies 1/5 = 1/2$$

which is not true, hence A and B are dependent events.

MY PROBLEM: I can't understand the reasoning. Practically speaking, a roll of Green die and roll of Red die are completely independent. One can not not affect the other. If Green die got 3 e.g. it can't affect the Red die to have or not to have 3. Then how they are dependent like author proved ?

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  • $\begingroup$ "B is an event where Green die is larger than event A." doesn't make sense. The event in A is "the red die is greater than 4" ... If the green die is showing a 5, say, how can you compare that with event A? one outcome is a number, the other is something that might happen; it's not a number. Do you you mean to compare the value on the green die to the value on the red die? $\endgroup$ – Glen_b Nov 18 '18 at 6:13
  • $\begingroup$ I edited the post to be clearer. I am comparing numbers on the dice. Event B is when number on green die is larger than the number in event A (red die $\ge 4$) $\endgroup$ – Arnuld Nov 18 '18 at 7:03
  • $\begingroup$ Your present definitions don't match the probabilities in your question. If B is the event "the number on the green die is greater than the number on the red die" that would have P(B) = 15/36. $\endgroup$ – Glen_b Nov 19 '18 at 3:15
  • $\begingroup$ Honestly I don't get your question ;( $\endgroup$ – Arnuld Nov 19 '18 at 4:09
  • $\begingroup$ Your question asserts a value for P(B), it says that it's 15/36, right? $\endgroup$ – Glen_b Nov 19 '18 at 4:33

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