Probability with given probability problem

[kenny1999]

Homework Statement

In order to be employed by a particular company, the interviewee has to attend one of two examination A and B and pass the exam. According to past experience, probability of passing examination A is 2/5 and that of B is 1/4. The ratio of interviewee attending examination A and B is 3:2

Given that the interviewee is employed by the company, what is the probability of he passing examination A?

Homework Equations

The Attempt at a Solution

No ideas at all. I always fear "given"

 

[Infinitum]

Hi kenny!

Start with, what is the probability that he will be employed by the company? Call this probability $P(I)$

Then think, with what probability in that total, are his chances of passing A?

Mathematically speaking, you are trying to find out what $P(A|I)$ is.

 

[Ray Vickson]

Homework Statement

In order to be employed by a particular company, the interviewee has to attend one of two examination A and B and pass the exam. According to past experience, probability of passing examination A is 2/5 and that of B is 1/4. The ratio of interviewee attending examination A and B is 3:2

Given that the interviewee is employed by the company, what is the probability of he passing examination A?

Homework Equations

The Attempt at a Solution

No ideas at all. I always fear "given"

Sometimes the best way to deal with this type of problem is to imagine a large number of candidates, say 5,000. How many of these write exam A? How many write exam B? Of those who write A, how many pass? Ditto for those who write B? Put all these numbers into a table.

So, the probability of passing is the number who pass A plus the number who pass B, all divided by 5000. You want a _conditional_ probability P(wrote A|pass). Can you see how to get that in terms of all the tabular entries?

Another way some people prefer is to draw a "tree diagram". In the first fork of the tree we have two branches "write A" and "write B". At the end of each of these two branches we have forks with additional branches "pass" or "fail". You can attach probabilities to each branch, then multiply them to get the probability at each of the four "tips".

A third way (which just formalizes the other two) is to use Bayes formulas; these are found in any textbook on the subject.

RGV

 

[kenny1999]

Sometimes the best way to deal with this type of problem is to imagine a large number of candidates, say 5,000. How many of these write exam A? How many write exam B? Of those who write A, how many pass? Ditto for those who write B? Put all these numbers into a table.

So, the probability of passing is the number who pass A plus the number who pass B, all divided by 5000. You want a _conditional_ probability P(wrote A|pass). Can you see how to get that in terms of all the tabular entries?

Another way some people prefer is to draw a "tree diagram". In the first fork of the tree we have two branches "write A" and "write B". At the end of each of these two branches we have forks with additional branches "pass" or "fail". You can attach probabilities to each branch, then multiply them to get the probability at each of the four "tips".

A third way (which just formalizes the other two) is to use Bayes formulas; these are found in any textbook on the subject.

RGV

sorry for late reply, i was not able to use cmoputer for a few days. two further questions about this. Thanks for replies.

1. In your first method.

Do you mean that I first caluclate the probability of P(Pass), then the resulting solution is = P(A | Pass) = P( A and Pass) / P ( Pass) ??

Anymore thing to take care of? Thank you

2. without using the concept of Bayes formula, is it the second method the best way to start with such kind of problems?

 

[HallsofIvy]

Yes, the best way to do problems like this is exactly what Ray Vinson said! Have you tried at all? Suppose there are 5000 applicants. Since the ratio of "A to B" is 3: 2, 3/5, or 3000, take exam A, 2/5, or 2000 take exam B. Since 2/5 pass exam A, 2/5 of 3000, or 1200, pass exam A. Since 1/4 pass exam B, 1/4 of 2000, or 500 pass exam B.

So how many people passed one of the exams (and so were hired)? Of those how many passed exam A?