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I understand well the concept of independence if it is rolling a balance die twice. of course, the first roll and the second roll are independent, because the first roll does not impact the second roll.

A:{ 6 appears in the first roll } P(A)= 1/6
B:{ 6 appears in the second roll } P(B)= 1/6
A ∩ B: {both rolls show 6} P(A ∩ B) = 1/36
P(A ∩ B)= P(A)P(B)

Now, I am rolling the die once. Can I say A and B are independent in the following case:

A = {1, 2}; P(A) = 2/6
B = {2, 3, 4} ; P(B) = 3/6
A∩B = {2}; P(A∩B) = 1/6   
P(A∩B) = P(A)P(B)

How to understand the concept of independance that event A = {1, 2} does not affect event B = {2, 3, 4} . I am confused.

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    How are the events A and B are definded ? – 2017-02-05
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    I edited my question. – 2017-02-05
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    And I have posted an anwser. – 2017-02-05

2 Answers 2

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In a way it is of course just coincidence that it works out that way. Imagine if you extend $A$ to $\{1,2,5\}$. Then it no longer is the same.

But, it still is true that the chance of $A$ happening is the same whether $B$ happens or not, since if $B$ happens (so we rolled a 2,3, or 4), then we have a 1/3 chance that $A$ happens (that we got the 2 out of those three), and if $B$ does not happen (so we got a 1,5,6) we again have a 1/3 chance.

So: $P(A|B)=P(A)$ (and hence $P(A \cap B) = P(A|B)*P(B)=P(A)*P(B)$) ... So while the presence of $B$ will effect the possible ways $A$ can happen, the chance of $A$ happening remains the same... and in that sense they are still said to be independent.

But you are right to be confused: the way in which these two events are independent is quite different from the way in which the outcomes of two dies are independent: wih two dice, the chance of getting a 6 is 1 out of six, no matter what the other die comes up with. (And this is how we typically think about independence). But with your example we go from 2 out of 6 to 1 out of 3 .... So the probability stays the same, even as the possible outcomes are effected!

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    Do you mean if P(A∩B)=P(A)∗P(B) , A and B are not always independent. or As long as P(A∩B)=P(A)∗P(B), A and B are independent although it might not be logical in English. – 2017-02-05
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    Well, the definition of Independence is P(A∩B)=P(A)∗P(B). meaning as long as this condition is met, the 2 events are independent (https://en.wikipedia.org/wiki/Independence_(probability_theory)) – 2017-02-05
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    @DiWang Good point! Ah! I see what is going on ... Let me change my answer... – 2017-02-05
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    A extend to {1,2,5}, B = {2, 3, 4} A∩B ={2}. they are still independent. – 2017-02-05
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    @DiWang No. With that, $P(A) =P(B)=\frac{1}{2}$, so $P(A)*P(B)=\frac{1}{4}$, but $P(A\cap B)=\frac{1}{6}$. – 2017-02-05
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    you are right ! – 2017-02-05
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    @DiWang please read my updated answer ... – 2017-02-05
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Let´s take the example of Bram28 first.

$A=\{1,2,5\}$ and $B=\{2,3,4\}$

Consequently the corresponding probabilities are $P(A)=\frac{1}{2}$ and $P(B)=\frac{1}{2}$. For independency $P(A| B)=P(A)$ must be true. If you notice that Event $B$ has occurred then the probability that event $A$ has occured is $P(A|B)=\frac{1}{3}$. Thus $P(A| B)\neq P(A)$. Or the other way round. If you notice that event $A$ has occurred then the probability that event $B$ has occured is $P(B|A)=\frac{1}{3}$.

Thus the events are not independent and $P(A\cap B)=\frac16$

Now have a quick look at the example of your question.

A = {1, 2}; P(A) = 2/6
B = {2, 3, 4} ; P(B) = 3/6

$P(A|B)=\frac{1}{3}$ Thus $P(A|B)=P(A)$. Here the events $A$ and $B$ are independent as you have already investigated. If you notice that event $B$ has occurred then the probability that event $A$ has occured is the same (unconditional) probability that event $A$ occcur.