Question

In four schools B₁, B₂, B₃ and B₄ the percentage of girl students is 12, 20, 13 and 17 respectively.From a school selected at random, one student is picked up at random and it is found that the student is a girl. The probability that the girl selected from school B₂ is

(a) \(\frac{6}{31}\)

(b) \(\frac{10}{31}\)

(c) \(\frac{13}{62}\)

(d) \(\frac{17}{62}\)

Answer 1

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}

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

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