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Note : $P(A)$ denotes the probability of an event.

Okay, so we say that events $A$ and $B$ are independent iff :

$$ P(A|B)=\frac{P(A∩B)}{P(B)} $$

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

In finding whether $A$ and $B$ are independent or not, we use this formula and check out whether $P(A|B)=P(A)$ or not. But how do i check independence when let's assume event $A$ is the number x we get when we roll a dice, and event $B$ is the head we get when we toss a coin. In this case how would we find $P(A∩B)$ or $A∩B$ in general?

Or do we conclude that since $A$ and $B$ are subsets of a different set, we automatically say that they are independent?

Help me out please.

Edit : try to unproven this for me :

You said that if two events come from different sets, they must be independent. Let's assume event $A$ and event $B$ come from different sets, that would mean that $A∩B=Ø$ Since intersection of our events is an empty set $P(A∩B)=0$. If we assume that event $A$ is getting number x when we roll a dice, $P(A)=\frac{1}{6}≠P(A∩B)$ thus $A$ is independent?

Here mathematically it looks like event $A$ is independent

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  • $\begingroup$ Please properly define $B$ as either head or tail, and define $A$ as an event instead of a random variable. $\endgroup$ – LinAlg Oct 12 '16 at 23:05
  • $\begingroup$ Edited, is it clear now? $\endgroup$ – Xenidia Oct 12 '16 at 23:13
  • $\begingroup$ An event is not a number. What is $P(A)$? $\endgroup$ – LinAlg Oct 12 '16 at 23:17
  • $\begingroup$ The definition of $P(A)$ is actually important, since we can use it to define the relevant part of the outcome space. $\endgroup$ – LinAlg Oct 12 '16 at 23:20
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But how do i check independence when let's assume event $A$ is the number $x$ we get when we roll a dice, and event $B$ is the head we get when we toss a coin. In this case how would we find $P(A∩B)$ or $A∩B$ in general?

Basically, an experiment can be described by a sample space consisting of a set of outcomes that describe all possible results, and the probability measure associated with them.

If the experiment consists of rolling a dice and flipping a coin, then an appropriate sample space is the set of Cartesian pairs of dice and coin results.

$$\left\{\begin{array}{c}\bbox[yellow,1pt]{(1,H)},&\bbox[lime,1pt]{(2,H)},&\bbox[yellow,1pt]{(3,H)},&\bbox[yellow,1pt]{(4,H)},&\bbox[yellow,1pt]{(5,H)},&\bbox[yellow,1pt]{(6,H)},\\(1,T),&,\bbox[cyan,1pt]{(2,T)},&(3,T),&(4,T),&(5,T),&(6,T)\end{array}\right\}$$

As to the probability measure: if the coin and dice are unbiased and don't influence each other, then all 12 outcomes are equally probable; though other experiments may have biases.   The intersection of the event of "coin result heads" (in $\bbox[yellow,1pt]{\text{yellow}}$ and $\bbox[lime,1pt]{\text{lime}}$) and the event of "dice result 2" ($\bbox[cyan,1pt]{\text{cyan}}$ and $\bbox[lime,1pt]{\text{lime}}$)) is the outcomes where both events occur (obviously $\bbox[lime,1pt]{\text{lime}}$).

So $\mathsf P(A\cap B)=\tfrac 1{12}= \tfrac{6}{12}\cdot\tfrac 2 {12} = \mathsf P(A)\cdot\mathsf P(B)$ as was anticipated from the intuition that the coin and dice don't influence each other.

tl;dr Just ensure that whatever notion of sample space and probability measure you use properly describes all the outcomes you can observe (or rather all those which you are interested in measuring).

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First, note that this condition is equivalent $P(A\cap B)=P(A)\cdot P(B)$.

Independence intuitively would mean something like 'orthogonality' in the space of 'possible states'.

For this concrete example, draw the probability space as a 1x1 square, and split it into 6 parts horizontally (one slice is event $A$) and 2 parts vertically (one part is event $B$).

More generally, assuming we can represent the probabilities within the unit square $[0,1]\times[0,1]$ such that $A=A_h\times[0,1]$ and $B=[0,1]\times B_v$ for some horizontal segment $A_h$ of length $a$ and vertical segment $B_v$ of length $b$, then we have $$P(A\cap B)= a\cdot b$$ because that's a rectangle of side $a$ and $b$.

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  • $\begingroup$ I see that, But I'm looking more of a verbal answer here. Since this is an statistics course which requires a verbal answer (introductory.) $\endgroup$ – Xenidia Oct 12 '16 at 23:20
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When determining if two events are independent, I usually go with my intuition, and you've stated it very well: if the events come from different sets, then they must be independent. Once I've decided on my intuition, then I prove it mathematically, as @Berci wrote: $P(A\cap B)=P(A)\cdot P(B)$

Note that not all independent events have different sets. For example, two flips of the same coin are independent, since one flip does not influence the other. You can prove this with the formula above, though some people may believe otherwise (their intuition is flawed). I guess what I'm saying is, use intuition if you can, and express it verbally as required in your course; but good courses will require to back up the verbal explanation with at least a minimum of math, which you now have at your disposal.

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  • $\begingroup$ maybe elaborate on how you compute the left hand side $\endgroup$ – LinAlg Oct 12 '16 at 23:41
  • $\begingroup$ You said that if two events come from different sets, they must be independent. Let's assume event $A$ and event $B$ come from different sets, that would mean that $A∩B=Ø$ Since intersection of our events is an empty set $P(A∩B)=0$. If we assume that event $A$ is getting number x when we roll a dice, $P(A)=\frac{1}{6}≠P(A∩B)$ thus $A$ is independent? Here mathematically it looks like event $A$ is independent $\endgroup$ – Xenidia Oct 12 '16 at 23:53
  • $\begingroup$ $A$ and $B$ are both from the same event space. As long as $A$ and $B$ are not mutually exclusive, $A\cap B \neq \emptyset$. $\endgroup$ – LinAlg Oct 13 '16 at 0:02
  • $\begingroup$ One does not say "event $A$ is independent." On says "events $A$ and $B$ are independent." See? One event cannot be independent by itself; two (or more) events are dependent or independent together. $\endgroup$ – scott Oct 13 '16 at 15:43
  • $\begingroup$ The notation $P(A\cap B)$ means "the probability of those two events occurring together." So if $A$ is rolling a six and $B$ is flipping a head, then $P(A\cap B)=\frac{1}{6}\cdot\frac{1}{2}=\frac{1}{12}$. This is the same as $P(A)\cdot P(B)$, so we know the events are independent. $\endgroup$ – scott Oct 13 '16 at 15:44

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