Question

Question: Suppose that \(E, {F_1},{F_2}\,and {F_3}\)are events from a sample space S and that \({F_1},{F_2}\,and {F_3}\) are pair wise disjoint and their union is S. Find \(p\left( {\frac{{{F_2}}}{E}} \right)\)if \(p\left( {\frac{E}{{{F_1}}}} \right) = \frac{2}{7},p\left( {\frac{E}{{{F_2}}}} \right) = \frac{3}{8},p\left( {\frac{E}{{{F_3}}}} \right) = \frac{1}{2},p\left( {{F_1}} \right) = \frac{1}{6},p\left( {{F_2}} \right) = \frac{1}{2}\) and \(p\left( {{F_3}} \right) = \frac{1}{3}\)

Short answer

Answer:

\(p\left( {\frac{{{F_2}}}{E}} \right) = 0.4018\)

Step-by-step solution

Given data

\(p\left( {\frac{E}{{{F_1}}}} \right) = \frac{2}{7},p\left( {\frac{E}{{{F_2}}}} \right) = \frac{3}{8},p\left( {\frac{E}{{{F_3}}}} \right) = \frac{1}{2},p\left( {{F_1}} \right) = \frac{1}{6},p\left( {{F_2}} \right) = \frac{1}{2}\)and\(p\left( {{F_3}} \right) = \frac{1}{3}\)

Formula used   

\(p\left( {\frac{E}{{{E_1}}}} \right) = \frac{{p\left( {\frac{{{E_1}}}{E}} \right)p\left( E \right)}}{{p\left( {\frac{{{E_1}}}{E}} \right)p\left( E \right) + p\left( {\frac{{{E_2}}}{F}} \right)p\left( F \right)}}\)

Calculating  

\(p\left( {\frac{{{F_2}}}{E}} \right) = \frac{{p\left( {\frac{E}{{{F_2}}}} \right)p\left( {{F_2}} \right)}}{{p\left( {\frac{E}{{{F_1}}}} \right)p\left( {{F_1}} \right) + p\left( {\frac{E}{{{F_2}}}} \right)p\left( {{F_2}} \right) + p\left( {\frac{E}{{{F_3}}}} \right)p\left( {{F_3}} \right)}}\)

\( = \frac{{\left( {\frac{3}{8}} \right)\left( {\frac{1}{2}} \right)}}{{\left( {\frac{2}{7}} \right)\left( {\frac{1}{6}} \right) + \left( {\frac{3}{8}} \right)\left( {\frac{1}{2}} \right) + \left( {\frac{1}{6}} \right)\left( {\frac{1}{3}} \right)}}\)

\( = \frac{7}{{15}}\)

\( = 0.4018\)

\(p\left( {\frac{{{F_2}}}{E}} \right) = 0.4018\)