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Q.State and prove Bayes' theorem.

Andhra Pradesh BieapBIEAP Intermediate Board 2020Subjective· 7mImportance★★★★★
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Bayes' theorem "reverses" a conditional probability, using the Total Probability Theorem to rewrite P(A)P(A) as a sum over a partition of the sample space.

Statement. Let E1,E2,…,EnE_1,E_2,\ldots,E_n be mutually exclusive and exhaustive events (a partition of the sample space SS) with P(Ei)>0P(E_i)>0 for every ii, and let AA be any event with P(A)>0P(A)>0. Then for each ii:

P(Ei/A)=P(Ei) P(A/Ei)∑j=1nP(Ej) P(A/Ej)P(E_i/A) = \frac{P(E_i)\,P(A/E_i)}{\displaystyle\sum_{j=1}^n P(E_j)\,P(A/E_j)}

Proof. Since E1,…,EnE_1,\ldots,E_n partition SS, any event AA can be split as:

A=A∩S=A∩(E1∪E2∪⋯∪En)=(A∩E1)∪(A∩E2)∪⋯∪(A∩En)A = A\cap S = A\cap(E_1\cup E_2\cup\cdots\cup E_n) = (A\cap E_1)\cup(A\cap E_2)\cup\cdots\cup(A\cap E_n)

Because the EiE_i are mutually exclusive, so are the pieces A∩EiA\cap E_i, so their probabilities simply add (Total Probability Theorem): …

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