Answer: (b) \(\frac{10}{31}\)
Let E1, E2, E3, E4 and A be the events defined as follows:
E1 = Event of selecting school B1
E2 = Event of selecting school B2
E3 = Event of selecting school B3
E4 = Event of selecting school B4
A = Event of selecting a girl.
Since there are four schools and each school has an equal chance of being chosen,
P(E1) = P(E2) = P(E3) = P(E4) = \(\frac{1}{4}\)
Now, P(Girl chosen is from school B1) = \(P\big(\frac{A}{E_1}\big)\) = \(\frac{12}{100}\)
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 B2
\( = \frac{P(E_2).P(A/E_2)}{P(E_1).P(A/E_1)+P(E_2).P(A/E_2)+P(E_3).P(A/E_3)+P(E_4).P(A/E_4)} \) Using Baye's theorem
\(\frac{\frac{1}{4}\times\frac{20}{100}}{\frac{1}{4}\times \frac{12}{100}+\frac{1}{4}\times \frac{20}{100}+\frac{1}{4}\times \frac{13}{100}+\frac{1}{4}\times\frac{17}{100}}\)
= \(\frac{\frac{1}{4}\times\frac{20}{100}}{\frac{1}{4}\times\frac{62}{100}}\) = \(\frac{20}{62}\) = \(\frac{10}{31}\)