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There is a question my friends and me have been going on in circles about. We're trying to understand how we can add up probabilities of mutually exclusive events, when there are conditional probabilities involved. Assume I have a die. If I roll this, I have an equal probability of getting anything from between 1 to 6. Now let me define 3 events $A =$ Rolling a 2, $B =$ rolling an event number, $X =$ rolling a prime. Now let's define $Y$ as $A\mid B$. $P(Y) = P(A|B) = 1/3$.


$$P(Y|X) P(X) + P(Y|X')(1-P(X)) = P(Y).$$

$P((A|B)|X) =$ probability of rolling a 2 if I rolled an even number ... if Ive rolled a prime. Working backwards, the primes are 2,3,5. The probability of rolling a 2 if the rolled an even number is 1. $P(X) = 1/2$.

$P((A|B)|X') =$ probability of rolling a 2 if I rolled an even number ... if I've rolled a non prime. Working backwards, the non-primes are 1,4,6. The probability of rolling a 2 if the rolled an even number is 0.

Hence $P(Y)$ should be $P((A|B)|X)\cdot 1/2 + P((A|B)|X')\cdot 1/2$, which should be $1/2$.

what are we missing?

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What do you mean by "define $Y$ as $A|B$"? $A|B$ is not an event, and there is no such thing as $P((A|B)|X)$, – Dilip Sarwate Apr 4 '12 at 20:46

What you're missing is that $A|B$ isn't an event. The usual notation for what you wrote as $P((A|B)|X)$ is $P(A|B,X)$, and $P(A|B,X)P(X)+P(A|B,X')(1-P(X))=P(A|B)$ isn't a law. The law $P(Y|X)P(X)+P(Y|X')(1-P(X))=P(Y)$ holds because the two probabilities being added are $P(Y\cap X)$ and $P(Y\cap X')$, respectively, and these are mutually exclusive events whose union is $P(Y)$, whereas there's no such interpretation for $P(A|B,X)P(X)$ and $P(A|B,X')(1-P(X))$, since there's no such thing as $P((A|B)\cap X)$.

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aha, so if I ask the question what's the probability of A given B, given X, what I'm calculating is P(A|B ∩ X). So does this mean I will not be able to sum up P(A|B ∩ X) and P(A|B ∩ X`) in any meaningful manner? – Vishnu Iyengar Apr 5 '12 at 5:09
@Vishnu: Well, the union of the disjoint events $A\cap B\cap X$ and $A\cap B\cap X'$ is $A\cap B$, so you have $P(A|B\cap X)P(B\cap X)+P(A|B\cap X')P(B\cap X')=P(A\cap B)$. – joriki Apr 5 '12 at 5:52

You're really seriously confused about the meaning of probabilistic notation. The notation $P(A \mid B)$ does not mean the probability of something called "$A\mid B$". Rather, it means the conditional probability given $B$, of something called $A$.

The notation $P((A\mid B)\mid X)$ is not correct notation no matter what it's intended to refer to.

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