Help with this difficult Probability Problem please

[JonJones]

Homework Statement

A student is studying for an exam, and a couple of weeks before the exam, the instructor gives him a list of all possible questions that may appear on the exam. Let this number of questions be W. On the exam, the instructor selects from these some questions to put on the exam. Let this number of questions be X.

Then, the instructor gives the student a choice from this number of X questions, of how many he must answer. Let this number of questions be Y. Before the exam, the student studies Z number of questions.

where W>X>Y

Find a function which yields (for every possible value of Z) the probability that this student will find himself dealing with at least one question he hasn't studied for.

Homework Equations

Minimum number of questions he must study in order to guarantee that he won't be dealing with at least one question he hasn't studied for is Z = W-(X-Y)
This wasn't given but I figured it out myself

The Attempt at a Solution

I thought it could help to look at it as a piece-wise function, where the probability would be 1 when Z<Y and 0 when Z≥W-(X-Y)
I guess I'm stuck on finding the function for Y< Z <W-(X-Y)


Thank you very much for your help

 

[tiny-tim]

Hi JonJones! Welcome to PF!

Find a function which yields (for every possible value of Z) the probability that this student will find himself dealing with at least one question he hasn't studied for.

This is one of those questions where it's easier to find the probability of it not happening (and then subtract from 1) …

so (in terms of X Y and Z) how would you describe that?

 

[adjacent]

Hi JonJones! Welcome to PF!


This is one of those questions where it's easier to find the probability of it not happening (and then subtract from 1) …

so (in terms of X Y and Z) how would you describe that?

Yes.As tiny-tim said,you will have to use 1-P(not happening) for questions where you find this:

at least

 

[JonJones]

As both of you said, I did try to find the probability of the negation and then doing 1 minus that. But I noticed that finding this probability is not anymore easier...

 

[tiny-tim]

As both of you said, I did try to find the probability of the negation …

show us how you described that negation (in terms of X Y and Z)

 

[JonJones]

show us how you described that negation (in terms of X Y and Z)

I really have no clue how to describe it. Never taken a probability course before.

 

[tiny-tim]

(just got up :zzz:)

ok, let's see how to rewrite this so that it's clearer

(It's an English thing, not a maths thing) …

A student is studying for an exam, and a couple of weeks before the exam, the instructor gives him a list of all possible questions that may appear on the exam. Let this number of questions be W. On the exam, the instructor selects from these some questions to put on the exam. Let this number of questions be X.

Then, the instructor gives the student a choice from this number of X questions, of how many he must answer. Let this number of questions be Y. Before the exam, the student studies Z number of questions.

where W>X>Y

Find a function which yields (for every possible value of Z) the probability that this student will find himself dealing with at least one question he hasn't studied for.

The probability of the negative of that is the probability that he can answer enough questions, ie that there are Y questions that he can answer.

He knows Z questions, and there are X questions altogether (W is irrelevant).

So, in English, and using the numbers X Y and Z, and words like "choose", how would you describe the probability that he can answer enough questions?

 

[JonJones]

(just got up :zzz:)

ok, let's see how to rewrite this so that it's clearer

(It's an English thing, not a maths thing) …The probability of the negative of that is the probability that he can answer enough questions, ie that there are Y questions that he can answer.

He knows Z questions, and there are X questions altogether (W is irrelevant).

So, in English, and using the numbers X Y and Z, and words like "choose", how would you describe the probability that he can answer enough questions?

I thought that W would be relevant because W choose X is the amount of all possible tests, and you would need to divide by this in the end to get the probability. Anyhow, it seems to me, that for Z≥Y (if Z<Y then the probability is 0) this probability would be:

Z choose Y
plus Z choose (Y+1)
...
plus Z choose Z

All over X choose Y?

 

[haruspex]

First, don't expect the answer to be necessarily a function in closed form. It could be a summation.
Second, I don't see how working with the negative form helps here.

Consider the event that there are exactly N questions on the test that the student has studied. You can partition the total set W into the X questions that are in the test and the W-X questions that are not. The questions that the student studies will be some from the X (how many?) and some from the W-X (how many?).
What is the probability of that event?