# Math Help - conditional probability 2

1. ## conditional probability 2

The probability a girl wears spectacles is 1/12 and the probability a student who wears spectacles is a girl is 1/9 . The probability that a student selected randomly is not a girl and does not wear spectacles is 3/4 . What is the probability that another person chosen randomly is a girl who wears spectacles ?

Let S=spectacles and G=girl

P(G n S)=1/12
P(G|S)=1/9
P(G' n S')=3/4

and the question is asking for P(G n S) again ??

2. Originally Posted by thereddevils
The probability a girl wears spectacles is 1/12 and the probability a student who wears spectacles is a girl is 1/9 . The probability that a student selected randomly is not a girl and does not wear spectacles is 3/4 . What is the probability that another person chosen randomly is a girl who wears spectacles ?

Let S=spectacles and G=girl

P(G n S)=1/12
P(G|S)=1/9
P(G' n S')=3/4

and the question is asking for P(G n S) again ??

No, it is asking for you to find the probability that a girl wearing spectacles is chosen GIVEN that the previous person was also a girl wearing glasses, so it's asking for:

$P(G \cap S | G \cap S)$

3. Originally Posted by Aryth
No, it is asking for you to find the probability that a girl wearing spectacles is chosen GIVEN that the previous person was also a girl wearing glasses, so it's asking for:

$P(G \cap S | G \cap S)$
thanks , but i don really understand , isn't that the same as G n S ??

(A n B) n (A n B) = A n B for example ?

4. Hi theredevils,

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

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

$\frac{1}{9}$ of the students who wear spectacles are girls.

$\frac{3}{4}$ 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

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

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

$\frac{3}{4}=\frac{C}{B+G}$

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

Solving this

$G=12F$

$3(B+G)=4C\ \Rightarrow\ B+G=\frac{4}{3}C$

$D=B-C\ \Rightarrow\ B-C=8F\ \Rightarrow\ C=B-8F$

$B+G=B+12F=\frac{4B-32F}{3}$

$3B+36F=4B-32F\ \Rightarrow\ B=68F$

$\frac{F}{B+G}=\frac{F}{68F+12F}=\frac{1}{80}$

5. Originally Posted by Archie Meade
Hi theredevils,

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

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

$\frac{1}{9}$ of the students who wear spectacles are girls.

$\frac{3}{4}$ 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

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

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

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

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

Solving this

$G=12F$

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

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

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

$\frac{F}{B+G}=\frac{F}{32F+12F}=\frac{1}{44}$
thanks a lot Archie, but the answer given by the book is 1/80 .

6. Aw cripes!

sorry theredevils,
i've made a misinterpretation then,
i'll endeavour to correct myself,
cheers.

7. Hi theredevils,

that error is corrected,

it was on line $\frac{3}{4}=\frac{C}{B+G}$