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What's the difference between P(A|B) and P(A|B=+ve) ? Are they the same or different ? Are the law of addition and multiplication applicable to both ?

Edit: Sorry, edited for clarity. I am talking about probability

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    What do you mean by "A|B=+ve"? Usually $A|B$ means that $A$ divides $B$, but that is a logical statement, not an expression that can meaningfully appear on one side of an equals sign.2011-10-24
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    If $A$ and $B$ are events (and $B$ has nonzero probability), then $P(A\mid B)$ is the probability that $A$ occurs given that $B$ occurs, that is $\frac{P(A \cap B)}{P(B)}$. I have no idea what $P(A \mid B=+\mathrm{ve})$ means.2011-10-24
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    Assume that the two possible values for Event B are +ve and -ve. P(A|B) is probability that A occurs given B and P(A|B=+ve) is the probability that A occurs given B is +ve. Are they the same ?2011-10-24
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    It still doesn't make sense. In $P(A|B)$, $B$ must be an _event_ (that is, a subset of the sample space), whereas in $P(A|B>0)$, $B$ must be a _random variable_ (that is, a function from the sample space to $\mathbb R$). The two expressions cannot both be meaningful at the same time!2011-10-24
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    Also, please don't abbreviate "positive" and "negative" as "+ve" and "-ve". Mathematics likes terse formulas, but it is _not_ text messaging. And, abbreviations aside, do not write nonsense like "$\ldots = \text{positive}$". The equals sign is not a universal abbreviation for "is"; its job is to construct _equations_ asserting that two expressions have the same value. It has no place if the things it connects are not expressions.2011-10-24
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    @HenningMakholm You're mistaken: $\Pr(A\mid B)$ can make perfect sense if $B$ is a random variable and $A$ is an event. Consider the meaning of $\Pr(A\mid B=b)$. That's a function of $b$; call it $g(b)$. Then $\Pr(A\mid B)$ means the random variable $g(B)$. However, I suspect what is meant by $B=+ve$ is that a patient tested positive for a disease, or something like that.2011-10-24
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    @Michael, I can see that is meaningful, but it surprises me that it would be a useful enough construction to be generally recognized.2011-10-24
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    @HenningMakholm It occurs in the law of total probability: If $A$ is an event and $B$ is a random variable, then $\mathbb{E}(\Pr(A\mid B)) = \Pr(A)$. In other words, the prior expected value of the posterior probability equals the prior probability.2011-10-25
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    @HenningMakholm .....and also in the law of total variance: $\operatorname{var}(X) = \operatorname{var}(\mathbb{E}(X\mid Y)) + \mathbb{E}(\operatorname{var}(X\mid Y))$.2011-10-25
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    @HenningMakholm .....and in the [Rao–Blackwell theorem](http://en.wikipedia.org/wiki/Rao-Blackwell_theorem).2011-10-27

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Maybe it's what are you looking for...

If $B=\{b_1,b_2\}$ or the event B have only two values and you call one of this, for example $b_1$ by True and the other, $b_2$, by False, so

$$P(A|B)=P(A|B=True)$$

because this is a convention for not always repeat $B=True$ every time you want to write it.

Example:

$Toothache=\{True,False\}$

$P(Caries|Toothache)=P(Caries|Toothache=True)$

by convention.