Changes

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in this case normalized to produce the equivalent statement about probabilities:

The terms of this equation can be normalized to produce the corresponding statement about probabilities:

{| align="center" cellspacing="6" width="90%"

{| align="center" cellspacing="6" width="90%"

Let's see if this checks out.

Let's see if this checks out.

Let <math>N\!</math> be the number of things in general, in Peirce's lingo, <math>N = [\mathbf{1}].</math>  On the assumption that <math>\mathrm{m}\!</math> and <math>\mathrm{b}\!</math> are associated with independent events, we get <math>[\mathrm{m,}\mathrm{b}] = \operatorname{P}(\mathrm{m}\mathrm{b}) \cdot N = \operatorname{P}(\mathrm{m})\operatorname{P}(\mathrm{b}) \cdot N = \operatorname{P}(\mathrm{m})[\mathrm{b}] = [\mathrm{m,}][\mathrm{b}],</math> so we have to interpret <math>[\mathrm{m,}]\!</math> = "the average number of men per things in general" as P(''m'') = the probability of a thing in general being a man.  Seems okay.

Let <math>N\!</math> be the number of things in general.  In terms of Peirce's "number of" function, then, we have the equation <math>[\mathbf{1}] = N.</math>  On the assumption that <math>\mathrm{m}\!</math> and <math>\mathrm{b}\!</math> are associated with independent events, we obtain the following sequence of equations:

 

<math>\begin{array}{lll}

[\mathrm{m,}\mathrm{b}]

& = &

\operatorname{P}(\mathrm{m}\mathrm{b}) N

\\[6pt]

& = &

\operatorname{P}(\mathrm{m})\operatorname{P}(\mathrm{b}) N

\\[6pt]

& = &

\operatorname{P}(\mathrm{m})[\mathrm{b}]

& = &

[\mathrm{m,}][\mathrm{b}]

\end{array}</math>

 

As a result, we have to interpret <math>[\mathrm{m,}]\!</math> = "the average number of men per things in general" as <math>\operatorname{P}(\mathrm{m})</math> = "the probability of a thing in general being a man".  This seems to make sense.

===Commentary Note 11.22===

===Commentary Note 11.22===