# Why doesn't this conditional probability equation hold

This is probably a very stupid question but I can't wrap my head around it.

$$P(B \cap A) = P(A \cap B) = P(B \mid A)\cdot P(A) + P(A \mid B)\cdot P(B)$$

Can someone explain intuitively why the above isn't true? What I am essentially saying here is that A and B both occurred is same as saying A occurred and then B occurred,following A OR B occurred and then A occurred, following B. So probability of A AND B happening = (Probability that A happened AND then probability that given A, B happened) OR (Probability that B happened AND then probability that given B, A happened)

Where am I going wrong?

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Do you see what goes wrong if $A=B$? – Sasha Mar 20 '12 at 21:27
Because most of the time $x\ne 2x$. – André Nicolas Mar 20 '12 at 21:35
I know that equation is incorrect.I wasn't able to grasp where my line of reasoning was going wrong to lead me to a wrong result.Got it! – prob Mar 20 '12 at 21:56

I get the impression that there may be a misunderstanding here in that you describe the terms you wanted to add as differing in the chronological order in which events A and B happened. That's not what conditional probabilities are about. Although in practice conditional probabilities often come into play when things occur in succession and we want to update our probabilities according to known outcomes, the conditional probabilities themselves have no temporal aspect. The two terms you wanted to add are the same because they are both the probability of A and B happening, expressed in two different ways in terms of conditional probabilities.

Imagine rolling two dice and recording both the numbers they show and the times at which they came to rest. To get the probability of rolling two $1$s, you can proceed in two different ways. You can ignore the times and just say that each die shows $1$ with probability $1/6$, for a joint probability of $1/36$. Or you can say that with probability $1/2$ die A comes to rest first, with probability $1/6$ it shows a $1$ and then die B comes to rest and shows a $1$ with probability $1/6$; and with probability $1/2$ die B comes to rest first, with probability $1/6$ it shows a $1$ and then die A comes to rest and shows a $1$ with probability $1/6$; so the total probability for one of these two ways of getting two $1$s to occur is $\frac12\cdot\frac16\cdot\frac16+\frac12\cdot\frac16\cdot\frac16=\frac12\cdot\frac1{36}+\frac12\cdot\frac1{36}=\frac1{36}$. The result is of course the same.

The point is that the way these things are usually treated, we don't care about chronological order, and the result can be interpreted as a sum over all possible chronological orders. If you do distinguish outcomes according to chronological order, you get an extra factor $1/n!$ for all chronological permutations of $n$ events. What you did was to distinguish according to chronological order but to use the probabilities as they're usually defined when one doesn't distinguish according to chronological order, but you didn't include the factor of $1/2$.

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You are counting too much. The probability of two events occuring is the probability that one event occurs, and then, given this event, what is the probability that the other one happens. Hence you are counting twice as many outcomes as you want.

Take for instance an easy example: Consider rolling a die. $A=\mbox{even number}$, $B=\mbox{multiple of 3}$. It is clear that $P(A\cap B)=\frac{1}{6}$.

Now given that the number is even, (your space is reduced to ${2,4,6}$) what is the probability that your number is a multiple of $3$, it is $\frac{1}{3}$, hence you have $\frac{3}{6}\frac{1}{3}=\frac{1}{6}$.

Given that the number is multiple of $3$, (your space is reduced to $\{3,6\})$ what is the probability that it is even, it is $\frac{1}{2}$. Hence, in this case you have $\frac{2}{6}\frac{1}{2}=\frac{1}{6}$.

If you add this probabilities you get $\frac{1}{3}$ which is double what the answer ought to be.

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I know that equation is incorrect.I wasn't able to grasp where my line of reasoning was going wrong to lead me to a wrong result. I wanted to add the probabilities in both those scenarios since they were both possible ways $A \bigcap B$ could happen.But that addition doesn't hold as those scenarios are not mutually exclusive, infact they are completely overlapping and hence leads to twice the actual probability. – prob Mar 20 '12 at 21:54

All these are great explanations. The thing that struck me as being out of place was the Plus sign. $A$ intersect $B$ is the same thing as $B$ intersect $A$. Using the conditional probability formula: $[p(A|B)p(B) = A \,\text{intersect} B]$ also $[P(B|A)P(A) = B \,\text{intersect} A]$. I consider it different ways of finding the same intersection probability. The choice of which formula depends on the available information. I believe changing the Plus sign to an equal sign would clear your confusion.

I am currently studying for the $P_1$ exam. If anyone sees fault with my explanation, please correct me.

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