# Why use A|BC and not ABC?

Currently I'm reading Probability Theory The Logic of Science and there it's not really clear for me what exactly $A|BC$ means. So I have two questions: Is this a general notation? Is $A|BC$ equal to $ABC$? If it is — that means it's just for the sake of order/consecution, like $A(BC)$ (but according to the book it's not so)?

• I don't have the book. Where is it first mentioned and what is the context (i.e., the first passage in which it appears)? – Sean Roberson Jun 13 '18 at 16:27
• In the context of probability this almost certainly is in reference to conditional probability, noting that your author apparently uses shorthand. I would have written as $Pr(A\mid B\cap C)$ while some authors replace $\cap$ with $,$ but yours apparently leaves it out entirely. – JMoravitz Jun 13 '18 at 16:41

In a probability theory course $x|y$ usually means "x given y". Elsewhere $x|y$ might mean "x divides y" or "the set of x such that y".

$x|y$, or "x given y" might come up in context as follows: Let

$x =$ Your car runs well
$y =$ Your car is on fire

Then $P(x|y) \leftrightarrow$ the probability that your car runs well given that your car is on fire.
$\big(P(x|y)$ is at or trending towards zero$\big)$

This notation will be important when studying conditional probability (for example, Bayes' Theorem or the Law of Total Probability)

$A|BC$ does not mean much

but the conditional probability $P(A \mid B,C)$ does, perhaps being read as the probability that event $A$ occurs given that the events $B$ and $C$ both occur;

This is not the same as the joint probability $P(A , B,C)$ which is the probability that all three events occur

As an example, suppose you throw three standard dice, and the events are:

• $A$ all three dice show the same value
• $B$ the sum of the three values is greater than or equal to $16$
• $C$ the sum of the three values is even

There are $6^3=216$ equally probable ways of throwing the dice, of which $6$ satisfy $A$, $10$ satisfy $B$, $108$ satisfy $C$, $7$ satisfy $B$ and $C$, and $1$ satisfies all three of $A$ $B$ and $C$.

So $P(A, B,C) = \frac{1}{216}$ but $P(A \mid B,C) = \frac{1}{7}$, rather different

You can show the relationship between the conditional probability and the joint probability with $$P(A \mid B,C)\, P(B,C) = P(A, B, C)$$

• It would be 6^3=216, not 6^2=216. – JustinCB Jun 13 '18 at 19:28
• @JustinC.B. - indeed - thank you – Henry Jun 13 '18 at 22:56