Prove Bayes Theorem

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Introduction to prove bayes theorem:-
Statement:-
                   Suppose B1,B2,B3.........................Bn are mutually exclusive and exhaustive events of a sample space S. such that       P ( Bi ) > 0, i = 1,2,................n and A is any arbitrary event of S such that P ( A ) > 0 and A `sube` `uu` Bi for i = 1 to n, Then the conditional Probability of Bi given P ( A ) for i = 1,2,.............n is equal to P ( Bi / A ) = ( P ( A / Bi ) P ( Bi ) ) / ( `sum_(i=1)^n` P ( ( A / Bi ) P ( Bi ) ).

Proof of Bayes Theorem
Given that B1,B2,B3.........................Bn are mutually exclusive and exhaustive events of a sample space S.
S = union of Bi and Bi `nn` Bj = `O/`  for all i `!=`j
A = sube`union of Bi for i = 1 to n
A = A  `nn` ( `uu` Bi ) for i = 1,2 .........n
A = A `nn` ( B1 `uu` B2 `uu` ..................................Bn )
A = ( A `nn` B1 ) `uu` ( A `nn` B2) `uu` .............................( A `nn` Bn)

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Take probability on both sides 
P ( A ) = P [ ( A `nn` B1 ) `uu` ( A `nn` B2) `uu` .............................( A `nn` Bn) ]
P ( A ) = P ( A `nn` B1 ) `uu` P ( A `nn` B2) `uu` .............................P ( A `nn` Bn)
P ( A ) = `sum_(i=1)^n` P ( A `nn` Bi )
P ( A ) = `sum_(i=1)^n` P ( Bi ) P ( A / Bi )            [`:.` P ( A `uu` B ) = P ( B ) P( A / B ) 
P ( A ) = `sum_(i=1)^n` P ( Bi ) P ( A / Bi ) --------------------- Equation 1

Conditional Probability Definition - Bayes Theorem
If A is any arbitrary event in a sample space and P ( A ) > 0. The probability of B given A is defined as
P ( B / A ) = P ( B `nn` A ) / P ( A )

This is called the conditional probability.
By the conditional probability of Bi / A is
P ( Bi / A ) = P ( Bi `nn` A ) / P ( A )
P ( Bi / A ) = { P ( Bi ) P ( A / Bi ) } / P ( A )
By equation 1 we can write the P (  A ) value
P ( Bi / A ) = { P ( Bi ) P ( A / Bi ) } / `sum_(i=1)^n` { P ( Bi ) P ( A / Bi ) }
Hence Proved

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