# In four schools B1, B2, B3 and B4 the percentage of girl students is 12, 20, 13 and 17 respectively.

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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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