**Answer: (b) **_{\(\frac{10}{31}\)}

**Let E**_{1}, E_{2}, E_{3}, E_{4} and A be the events defined as follows:

E_{1} = Event of selecting school B_{1 }

E_{2} = Event of selecting school B_{2}

E_{3} = Event of selecting school B_{3}

E_{4} = Event of selecting school B_{4 }

A = Event of selecting a girl.

Since there are four schools and each school has an equal chance of being chosen,

P(E_{1}) = P(E_{2}) = P(E_{3}) = P(E_{4}) = \(\frac{1}{4}\)

Now, P(Girl chosen is from school B_{1}) = \(P\big(\frac{A}{E_1}\big)\) = \(\frac{12}{100}\)

Similarly \(P\big(\frac{A}{E_2}\big)\) =\(\frac{20}{100}\)

\(P\big(\frac{A}{E_3}\big)\) =\(\frac{13}{100} \)

\(P\big(\frac{A}{E_4}\big)\)= \(\frac{17}{100}\)

∴ P(Girl chosen is from school B_{2}

_{\( = \frac{P(E_2).P(A/E_2)}{P(E_1).P(A/E_1)+P(E_2).P(A/E_2)+P(E_3).P(A/E_3)+P(E_4).P(A/E_4)} \) Using Baye's theorem}

_{\(\frac{\frac{1}{4}\times\frac{20}{100}}{\frac{1}{4}\times \frac{12}{100}+\frac{1}{4}\times \frac{20}{100}+\frac{1}{4}\times \frac{13}{100}+\frac{1}{4}\times\frac{17}{100}}\)}

_{= \(\frac{\frac{1}{4}\times\frac{20}{100}}{\frac{1}{4}\times\frac{62}{100}}\) = \(\frac{20}{62}\) = \(\frac{10}{31}\)}