1. ## Probability Problem

Hey all,

Haven't ever been one for probability.

There are 150 options for students to choose from. Student 1 picks an option and then it is removed from the pool, Student 2 then picks and follows suit. Each option is given equal weight. What is the probability of any student obtaining their choice?

I got stuck generalizing this:

P of first student getting his or her choice: P(1) = 1
P of second student getting his or her choice: P(2) = 1 - probability of first student taking their choice = 1 - 1/150 = 149/150

at this point i wonder if i'm already heading down the wrong path... shouldn't it be more like P(2) = 148/149?

P(3) = 1 - probability of first and second student taking their choice = 1 - (1/150) - (1/149)

I'm afraid to go any further without first understanding what I'm doing... Initially I thought it would turn out to be something like the probability of drawing two aces from a deck 4|52 * 3|51 , but I'm slamming my head against the wall now. Any help would be appreciated.

2. ## Re: Probability Problem

If I'm understanding your problem, then there's no decrease in your denominator.
$\frac{150}{150}, \frac{149}{150}, \frac{148}{150}, \frac{147}{150}, \frac{146}{150}... \frac{0}{150}.$

3. ## Re: Probability Problem

Maybe. I'm not sure I described it correctly. If there are 150 options, and each option can only be picked once (kind of like removing a marble from a pile) won't this decrease the denominator? Some of my initial thoughts were of the form

product over (150-k)/(151-k) from k = 1 to student number of interest

I'm still thinking along the lines of drawing two aces from a deck. If student 52 wants a certain option, then the 51 students before him have to not pick his option. The 51st student would have 100 options to decide from... am I making any sense?

4. ## Re: Probability Problem

Originally Posted by acls2012
There are 150 options for students to choose from. Student 1 picks an option and then it is removed from the pool, Student 2 then picks and follows suit. Each option is given equal weight. What is the probability of any student obtaining their choice?
I find this question so poorly worded.
In particular, what does "any student obtaining their choice" mean exactly?

Say the names of the students are in the hat. Each student picks a slip from the hat at random and keeps it.
Is it asking for example, What is the probability that John Adams draws his own name?

OR is it asking What is the probability that at least one student draws his/her own name?

I think it is meant to be the latter. But which is it?

5. ## Re: Probability Problem

I apologize for my obtuse wording. Perhaps I can restate my question. A class consists of 150 students. Each student is supposed to select a number 1 to 150. Beginning with the first student. After that particular student picks a number, the number is removed from the pool (analogous to your name and hat scenario). Assuming that each student makes a decision in their head about which number they would like, what is the probability that a given student will be able to select their number when it is their turn to select? The selection process itself is not dependent on chance. If John wants number 32 and 32 is available in the pot he will pick that number but its possible that another student before him also wanted that number, so it has already been removed

A concrete question might look like this: What is the probability of the 52nd student drawing the number he wants.

6. ## Re: Probability Problem

Originally Posted by acls2012
I apologize for my obtuse wording. Perhaps I can restate my question. A class consists of 150 students. Each student is supposed to select a number 1 to 150. Beginning with the first student. After that particular student picks a number, the number is removed from the pool (analogous to your name and hat scenario). Assuming that each student makes a decision in their head about which number they would like, what is the probability that a given student will be able to select their number when it is their turn to select? The selection process itself is not dependent on chance. If John wants number 32 and 32 is available in the pot he will pick that number but its possible that another student before him also wanted that number, so it has already been removed
Well the answer is $\frac{1}{150}$ that John gets his 32,

Here is why. Think of a roster of this class. Go down that list and randomly assign one of those 150 numbers to each name using all numbers exactly once( that models student choice). There are $150!$ ways to do that. There are $149!$ of those in which 32 will be next to John's entry in the roster. Divide, what do you get?

7. ## Re: Probability Problem

Thank you Plato.