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Q.State and prove Baye's theorem on probability.

Andhra Pradesh BieapBIEAP Intermediate Board 2023Subjective· 7mImportance★★★★★
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Bayes' theorem "reverses" conditional probability — from the probability of an effect given a cause, it recovers the probability of a cause given the observed effect; it follows directly from the multiplication rule plus the law of total probability.

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\mid A) = \frac{P(E_i)\,P(A\mid E_i)}{\displaystyle\sum_{j=1}^{n}P(E_j)\,P(A\mid E_j)}.

Proof. By the definition of conditional probability,

P(Ei∣A)=P(Ei∩A)P(A).P(E_i\mid A) = \frac{P(E_i\cap A)}{P(A)}.

By the multiplication rule, P(Ei∩A)=P(Ei) P(A∣Ei)P(E_i\cap A) = P(E_i)\,P(A\mid E_i), so

P(Ei∣A)=P(Ei) P(A∣Ei)P(A).(∗)P(E_i\mid A) = \frac{P(E_i)\,P(A\mid E_i)}{P(A)}. \quad(*)

Now find P(A)P(A) in terms of the EjE_j's. Since E1,…,EnE_1,\ldots,E_n are exhaustive and mutually exclusive, S=E1∪E2∪⋯∪EnS=E_1\cup E_2\cup\cdots\cup E_n, so

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

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