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

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Answer: (b) \(\frac{10}{31}\)

Let E1, E2, E3, E4 and A be the events defined as follows: 

E1 = Event of selecting school B

E2 = Event of selecting school B2 

E3 = Event of selecting school B3 

E4 = Event of selecting school B

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

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

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

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