Find today, very first, your proposition \(P\) enters just for the basic and the third of them site, and furthermore, the specifics out of these site is very easily safeguarded

Eventually, to determine the second end-that’s, you to definitely prior to our very own background knowledge as well as suggestion \(P\) its apt to be than simply not that Goodness does not occur-Rowe means only 1 extra assumption:

\[ \tag <5>\Pr(P \mid k) = [\Pr(\negt G\mid k)\times \Pr(P \mid \negt G \amp k)] + [\Pr(G\mid k)\times \Pr(P \mid G \amp k)] \]

\[ \tag <6>\Pr(P \mid k) = [\Pr(\negt G\mid k) \times 1] + [\Pr(G\mid k)\times \Pr(P \mid G \amp k)] \]

\tag <8>&\Pr(P \mid k) \\ \notag &= \Pr(\negt G\mid k) + [[1 – \Pr(\negt G \mid k)]\times \Pr(P \mid G \amp k)] \\ \notag &= \Pr(\negt G\mid k) + \Pr(P \mid G \amp k) – [\Pr(\negt G \mid k)\times \Pr(P \mid G \amp k)] \\ \end
\]
\tag <9>&\Pr(P \mid k) – \Pr(P \mid G \amp k) \\ \notag &= \Pr(\negt G\mid k) – [\Pr(\negt G \mid k)\times \Pr(P \mid G \amp k)] \\ \notag &= \Pr(\negt G\mid k)\times [1 – \Pr(P \mid G \amp k)] \end
\]

However in view from presumption (2) i’ve that \(\Pr(\negt Grams \mid k) \gt 0\), whilst in look at presumption (3) i have you to \(\Pr(P \mid Grams \amplifier k) \lt 1\), and thus 15 yД±llД±k yaЕџ farkД± iliЕџkileri you to \([1 – \Pr(P \mid G \amp k)] \gt 0\), so that it up coming comes after out of (9) one

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