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)?
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$\begingroup$ 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)? $\endgroup$ – Sean Roberson Jun 13 '18 at 16:27
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6$\begingroup$ 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. $\endgroup$ – JMoravitz Jun 13 '18 at 16:41
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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)
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$A|BC$ 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 $$P(A \mid B,C)\, P(B,C) = P(A, B, C)$$
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