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The updated chance of an event occurring after additional information is taken into account is known as posterior probability.

Given:

\(P(B) > 0\)

To proof:

\({\rm{P(A}}\mid {\rm{B) + P}}\left( {{{\rm{A}}'}\mid {\rm{B}}} \right){\rm{ = 1}}\)

Note: \({{\rm{A}}'}\)represents the complement of \({\rm{A}}\).

Proof: Definition Conditional probability:

\(P(B\mid A) = \frac{{P(A \cap B)}}{{P(A)}}\)

Using this definition, we then obtain:

\(\begin{aligned}P(A\mid B) &= \frac{{P(A \cap B)}}{{P(B)}}\\P(B\mid A) &= \frac{{P\left( {{A^\prime } \cap B} \right)}}{{P(B)}}\end{aligned}\)

The event \({A^\prime } \cap B\)is the part of event \({\rm{B}}\) that does not belong to event \({\rm{A}}\) :

\(P\left( {{A^\prime } \cap B} \right) = P(B) - P(A \cap B)\)

Combining the results, we then obtain:

\(\begin{aligned}P(A\mid B) + P\left( {{A^\prime }\mid B} \right) &= \frac{{P(A \cap B)}}{{P(B)}} + \frac{{P\left( {{A^\prime } \cap B} \right)}}{{P(B)}}\\ &= \frac{{P(A \cap B)}}{{P(B)}} + \frac{{P(B) - P(A \cap B)}}{{P(B)}}\\ &= \frac{{P(A \cap B) + P(B) - P(A \cap B)}}{{P(B)}}\\ &= \frac{{P(B)}}{{P(B)}} = 1\end{aligned}\)

Therefore, the solution is \(P(A\mid B) + P\left( {{A^\prime }\mid B} \right) = \frac{{P(B)}}{{P(B)}} = 1\)