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Q.State and prove Baye's Theorem.

Andhra Pradesh BieapBIEAP Intermediate Board 2019Subjective· 7mImportance★★★★★
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Bayes' theorem expresses the reverse conditional probability P(Ei∣A)P(E_i\mid A) in terms of the given (forward) conditionals P(A∣Ej)P(A\mid E_j), using the definition of conditional probability and the total-probability theorem.

Statement.

Let E1,E2,…,EnE_1,E_2,\dots,E_n be mutually exclusive and exhaustive events (a partition of the sample space SS) with P(Ei)>0P(E_i)>0 for each 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\mid A)=\dfrac{P(E_i)\,P(A\mid E_i)}{\displaystyle\sum_{j=1}^{n}P(E_j)\,P(A\mid E_j)}.

Proof.

Step 1 — Definition of conditional probability.

P(Ei∣A)=P(Ei∩A)P(A)P(E_i\mid A)=\dfrac{P(E_i\cap A)}{P(A)}, since P(A)>0P(A)>0.

Step 2 — Multiplication rule.

P(Ei∩A)=P(Ei) P(A∣Ei)P(E_i\cap A)=P(E_i)\,P(A\mid E_i).

Step 3 — Total probability of AA.

Since E1,…,EnE_1,\dots,E_n partition SS: 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), and since the EjE_j are mutually exclusive, so are the A∩EjA\cap E_j. Hence

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