1. ## probability

the probability a girl who 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. wht is the probability that a person chosen randomly is a girl who wear spectacles? answer provided1/80
2)among 10 000 candidATES who are applying to be admitted to teaching college, only 2200 are selected to sit for the written test. For evry 100 candidates who sit for the written test, 70 will be called for an oral interview. For every 10 candidates interviewed, 3 will successfully join the Teaching colleges.
a) what is the probability that an applicant will reach the stage of the interview?(0.154)
b) What is the probability that an applicant who is sitting for the written test suceeds in entering a TeachingCollege?(0.21)
c) What is the probability an applicant will succeed in entering a teaching college?(0.0462)
3)A dice is tossed 4 times. find the probability that we will obtain the number 4 exactly t.(25/256)

2. Draw a box. Put girl and not girl across the top. Put spectacles and not spectacles down the side. Fill in the probabilities given in the problem statement. Calcualte the rest.

3. did you mean set or box? -.- . pls briefly describe the box you mean and i will fill it . thx

4. I mean a box.

Have you ever played battleship or used a spreadsheet?

In Cell A1 put "Girl/Glasses Problem"
In Cell B1 put "Girl"
In Cell C1 put "Not Girl"
In Cell D1 put "Total"

A2 put "Spectacles"
B2 put a <== A value we don't know yet
C2 put b <== A value we don't know yet
D2 put c <== A value we don't know yet

A3 put "Not Spectacles"
B3 put d <== A value we don't know yet
C3 put e <== A value we don't know yet
D3 put f <== A value we don't know yet

A4 put "Totals"
B4 put g <== A value we don't know yet
C4 put h <== A value we don't know yet
D4 put i <== A value we don't know yet

That's just a setup. Now let's start thinking about it.

These should be obvious. Don't change any boxes, just think about htem.

a+b=c
d+e=f
g+h=i
a+d=g
b+e=h
c+f=i
Also, a+b+d+e=i

Now the facts.

"the probability a girl who wears spectacles is 1/12"

I think this means a = 1 and g = 12. Substitute those.

"the probability a student who wears spectacles is a girl is 1/9"

This means a = 1 (which thing we know already) and c = 9. Substitute those if you have not already.

"probability that a student selected randomly is not a girl and does not wear spectacles is 3/4."

This one is a little trickier. It means e = (3/4)i

It should be obvious now that d = 11 and b = 8

Almost the last thing. It should be obvious that h = 8 + (3/4)i and f = 11 + (3/4)i

If you have followed so far, you are almost done. Only a little algebra remains. Find 'i' 2 ways and see that you have a solid answer.

Very useful, a little box.

5. Originally Posted by qweiop90
did you mean set or box? -.- . pls briefly describe the box you mean and i will fill it . thx
The name of the box is a Karnaugh table. You will find many posts on these forums that explain it. Do a search of these forums.

6. Sorry but your first n 3rd question aint very clear.

However here is the solution for your second problem....
Out of 10,000 applicants 2200 will be goin to sit for written test..
N for evry 100 of those who r sittin in written test 70 will be called for an interview.
that means for 2200 apllicants 2200*70/100 =1540 will be called for the interview.
that means 1540 applicants will be selected.
Now this 1540 is to be divided by the total no. of outcomes i.e 10,000 n hence will get the ans of part a.

now as you know that 1540 applicants are to be called for the interview n out of every 10 of em 3 are to be selected... so total applicants who will be succeedin in enterin college is 1540*3/10 = 462..

so probability of applicants selected for written test goin to college is 462/2200 = .21

n probability of an applicant goin to college is 462/10,000 = .0462
hence ques solved.

7. Originally Posted by mr fantastic
The name of the box is a Karnaugh table. You will find many posts on these forums that explain it.
Really? I wonder why I never heard of it except in a previous post where you said that. "Karnaugh" it is, then. My apologies to the originator.

8. hello, your method about probability(ques 1 spectacles) r too complicated for me. In my syllabus , there is no such things exist . i am currently persue on Stpm. In our syllabus , only 3 events, conditional events,independents events,tree diagram are exist in our syllabus.

SO, do you have any idea on how to solve the probability (ques 1 spec) by using STPM methods?

9. Originally Posted by qweiop90
hello, your method about probability(ques 1 spectacles) r too complicated for me.
No it isn't. The description is rather complicated, but the actual method is rather simple. Did you put it in a spreadsheet? If you follow the instructions carefully, one step at a time, you will see it unfold. We're just putting the information in a rectangular grid and adding some labels and some totals. Only simple algebra is needed. The difficult part is in the translation of the given information.

In our syllabus , only 3 events, conditional events,independents events,tree diagram are exist in our syllabus.
It is very closely ralated to the tree diagram and to a discussion on conditional events. I think it is in your book, although perhaps not formally stated.

10. perhaps , i have never done this before. so, i have put my hardest effort to search abt the table. yet i stil couldnt understand. y dont you show me the plotting of the table. Since, it does not make any sense

11. Originally Posted by qweiop90
perhaps , i have never done this before. so, i have put my hardest effort to search abt the table. yet i stil couldnt understand. y dont you show me the plotting of the table. Since, it does not make any sense
Did you serach for and read any of the posts in these forums that use the idea? Eg. http://www.mathhelpforum.com/math-he...ease-help.html

12. yah, searched. yet i stil couldnt placed 4 variables correctly.

13. Originally Posted by qweiop90
yah, searched. yet i stil couldnt placed 4 variables correctly.
$\displaystyle \begin{tabular}{l | c | c | c} & G & G$\, '$& \\ \hline S & a & b & \\ \hline S$\, '$& c & 3/4 & \\ \hline & & & 1 \\ \end{tabular}$

Given:

$\displaystyle \Pr(S | G) = \frac{1}{12} \Rightarrow \frac{a}{a+c} = \frac{1}{12} \Rightarrow 11a = c$ .... (1)

$\displaystyle \Pr(G | S) = \frac{1}{9} \Rightarrow \frac{a}{a+b} = \frac{1}{9} \Rightarrow 8a = b$ .... (2)

$\displaystyle a + b + c = \frac{1}{4}$ .... (3)

Solve equations (1), (2) and (3) simultaneously for a.

$\displaystyle \Pr(G \cap S) = a$.