BAYE’S THEOREM

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Let E1, E2, E3, …………. En mutually exclusive and exhaustive events and A is the event occurs after E1, E2, E3, …………. En.

By multiplication theorem,

P\left( {A \cap \,{E_i}} \right) = P\left( A \right).P\left( {{E_i}/A} \right)

 \Rightarrow \,P\left( {{E_i}/A} \right)\, = \,\frac{{P\left( {A \cap \,{E_i}} \right)}}{{P\left( A \right)}} 

P\left( {{E_i}/A} \right)\, = \,\frac{{P\left( {{E_i}} \right).P\left( {A/{E_i}} \right)}}{{P\left( {{E_1}} \right).P\left( {A/{E_1}} \right) + P\left( {{E_2}} \right).P\left( {A/{E_2}} \right) + ........ + P\left( {{E_n}} \right).P\left( {A/{E_n}} \right)}} , where i = 1, 2, 3, … n

Example 1: A man is known to speak truth 3 out of 4 times. He throws a die and reports it is a six. Find the probability that it is actually a six.

Solution:

E1: Six comes up.

E2: Six does not comes up.

A: The man reports it is a six.

\begin{array}{l}
P\left( {{E_1}} \right) = \frac{1}{6}\\
P\left( {{E_2}} \right) = \frac{5}{6}
\end{array}

P\left( {A/{E_1}} \right) = {\rm{Probability}}\,{\rm{of}}\,{\rm{the}}\,{\rm{reports}}\,{\rm{it}}\,{\rm{is}}\,{\rm{six}}\,{\rm{and}}\,{\rm{it}}\,{\rm{is}}\,{\rm{actually}}\,{\rm{a}}\,{\rm{six}}

 = \,\frac{3}{4}

P\left( {A/{E_2}} \right)\, = \,{\rm{Probability}}\,\,{\rm{of}}\,{\rm{the}}\,{\rm{reports}}\,{\rm{it}}\,{\rm{is}}\,{\rm{a}}\,{\rm{six}}\,{\rm{but}}\,{\rm{six}}\,{\rm{not}}\,{\rm{comes}}\,{\rm{up}}

 = \,\frac{1}{4} 

\therefore By Baye’s theorem, required probability is:

P\left( {{E_1}/A} \right) = \,\frac{{P\left( {{E_1}} \right).P\left( {A/{E_1}} \right)}}{{P\left( {{E_1}} \right).P\left( {A/{E_1}} \right) + P\left( {{E_2}} \right).P\left( {A/{E_2}} \right)}}

 = \,\,\frac{{\frac{1}{6}\, \times \frac{3}{4}}}{{\frac{1}{6} \times \frac{3}{4} + \frac{5}{6} \times \frac{1}{4}}}

 = \,\frac{3}{{3 + 5}}

 = \,\frac{3}{8} 

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