Originally Posted by

**Archie Meade** Hi theredevils,

here's an alternative way, a bit long though.

$\displaystyle \frac{1}{12}$ of the all the girls wear spectacles.

$\displaystyle \frac{1}{9}$ of the students __who wear spectacles__ are girls.

$\displaystyle \frac{3}{4}$ of the students are boys who don't wear spectacles.

We are not directly given

...the fraction of all the students that are girls.

The word "another" may be offputting.

The following calculation finds the fraction of students that are girls who wear spectacles.

The question doesn't state that a person was chosen before another. It only gives another probability.

E....girls who don't wear spectacles

F....girls who do

C....boys who don't

D....boys who do

The total number of students are E+F+C+D=B+G

G=E+F

B=C+D

$\displaystyle \frac{1}{12}=\frac{F}{G}$

$\displaystyle \frac{1}{9}=\frac{F}{F+D}$

$\displaystyle \frac{3}{4}=\frac{C}{B}$

$\displaystyle P(random\ student\ =\ girl\ who\ wears\ spectacles)\ =\ \frac{F}{B+G}$

Solving this

$\displaystyle G=12F$

$\displaystyle 3B=4C\ \Rightarrow\ C=\frac{3}{4}B$

$\displaystyle D=B-C\ \Rightarrow\ D=\frac{B}{4}\ \Rightarrow\ B=4D$

$\displaystyle F+D=9F\ \Rightarrow\ D=8F\ \Rightarrow\ B=4(8F)=32F$

$\displaystyle \frac{F}{B+G}=\frac{F}{32F+12F}=\frac{1}{44}$