First success, sampling without replacement

[alman9898]

Homework Statement

A box contains 5 keys, only one of which will open a lock. Keys are randomly selected and tried, one at a time, until the lock is opened (keys that do not work are discarded before another is tried). Let Y be the number of the trial on which the lock is opened. What is the prob function for Y?

Homework Equations

{{{m \choose k} {{N-m} \choose {n-k}}}\over {N \choose n}}

The Attempt at a Solution

What I am trying to find...

P(Y = i) where i is the number of the trial where the "first" (and only) success occurs. This would be much easier if the trials were independent but since they are dependent, I'm having trouble translating the problem into a hypergeometric one. The explanations I"ve found provided online quickly jump to the math without explaining why they did what they did...

What I've tried is C(1,1) * C(4, i-1) / C(5, i) but I know this is wrong because I believe it should turn out that p(i) = 1/5 for all i (based off this http://docs.google.com/gview?a=v&q=cache:_nd2dQm5Me8J:https://www.cs.drexel.edu/classes/ProbStat/mcs311_Spring98/test2ans.pdf+%22a+box+contains+5+keys%22&hl=en&gl=us&sig=AFQjCNHgIMqwTOs-06Phz8C11i5iqKDTlA" I found online), but I can't determine how to proceed...I'm very confused...

Further update: it seems this formula probably describes the cdf rather than the pdf, but I don't know why...I would really appreciate a clear and concise explanation.

 

[LCKurtz]

Its obvious P(Y=1) = 1/5. To get P(Y=2) two things must happen. You must fail on the first draw and succeed on the second. So you need a conditional probability:

P(Y=2) = P(Y\ne 1)P(Y=2|Y\ne 1)

Can you figure that out and take it from there?

 

[alman9898]

So, what exactly did I find when I have

C(1,1)*C(4, i-1)/C(5,i) ... the conditional probability for Y = i ?

 

[mon176]

just a quick question, did you get your answer, because I am dealing with similar problem atm and can't find any help online with it, cheers

 

[LCKurtz]

Its obvious P(Y=1) = 1/5. To get P(Y=2) two things must happen. You must fail on the first draw and succeed on the second. So you need a conditional probability:

P(Y=2) = P(Y\ne 1)P(Y=2|Y\ne 1)

Can you figure that out and take it from there?

So, what exactly did I find when I have

C(1,1)*C(4, i-1)/C(5,i) ... the conditional probability for Y = i ?

Somehow I didn't see your answer earlier. I don't know what your forumula represents, but you ignored my question. The answer to my question will lead you to the solution method.

 

[Ray Vickson]

Homework Statement

A box contains 5 keys, only one of which will open a lock. Keys are randomly selected and tried, one at a time, until the lock is opened (keys that do not work are discarded before another is tried). Let Y be the number of the trial on which the lock is opened. What is the prob function for Y?

Homework Equations

{{{m \choose k} {{N-m} \choose {n-k}}}\over {N \choose n}}

The Attempt at a Solution

What I am trying to find...

P(Y = i) where i is the number of the trial where the "first" (and only) success occurs. This would be much easier if the trials were independent but since they are dependent, I'm having trouble translating the problem into a hypergeometric one. The explanations I"ve found provided online quickly jump to the math without explaining why they did what they did...

What I've tried is C(1,1) * C(4, i-1) / C(5, i) but I know this is wrong because I believe it should turn out that p(i) = 1/5 for all i (based off this http://docs.google.com/gview?a=v&q=cache:_nd2dQm5Me8J:https://www.cs.drexel.edu/classes/ProbStat/mcs311_Spring98/test2ans.pdf+%22a+box+contains+5+keys%22&hl=en&gl=us&sig=AFQjCNHgIMqwTOs-06Phz8C11i5iqKDTlA" I found online), but I can't determine how to proceed...I'm very confused...

Further update: it seems this formula probably describes the cdf rather than the pdf, but I don't know why...I would really appreciate a clear and concise explanation.

Your formula above is the answer to a question completely from the one you are asking about. Rather than writing down random formulas (that may, or may not be relevant), just sit down and consider the situation one key at a time. What is the probability that the first key works? What is the probability that the first key does not work but the second one does? After answering these two questions you should start to see the needed pattern.

RGV