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Choosing without replacement

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
We have a container that contains r balls that have numbers 1,...,r. We choose at random n of them without replacement. Let Y be the greatest and Z the smallest of the numbers of the balls we chose. Which are the probabilities P(Y<=y) and P(Z>=z)???

Thanks in advance!!!!
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,793
Re: Probability-Probability!choose without replacement!

We have a container that contains r balls that have numbers 1,...,r. We choose at random n of them without replacement. Let Y be the greatest and Z the smallest of the numbers of the balls we chose. Which are the probabilities P(Y<=y) and P(Z>=z)???

Thanks in advance!!!!
Hi evinda! ;)

Your problem looks a bit odd.
Is there perhaps something missing from the problem statement?

Anyway, if you choose n balls, then you always get n balls.
So the probability of n balls is one, and the probability to anything other than n balls is zero.
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Re: Probability-Probability!choose without replacement!

No,it is not something missing from the problem statement...And how can I find the possibilities P(Y<=y) and P(Z>=z)???? :confused:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,793
Re: Probability-Probability!choose without replacement!

No,it is not something missing from the problem statement...And how can I find the possibilities P(Y<=y) and P(Z>=z)???? :confused:
Oh wait! I misread. :eek:

To calculate $P(Y\le y)$ you need to know how many combinations there are with a highest number of at most y.
That is, in how many ways can you draw n balls from the set of the first y balls?

Can you say how many?

To get $P(Y\le y)$, you need to divide this by the total number of ways you can draw n balls from the total set of r balls.
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Re: Probability-Probability!choose without replacement!

So is the answer:p(Y<=y)= {y choose n}/{r choose n}??? And P(Z>=z)=1-P(Z<z)=1-{z choose n}/{r choose n} ???


Oh wait! I misread. :eek:

To calculate $P(Y\le y)$ you need to know how many combinations there are with a highest number of at most y.
That is, in how many ways can you draw n balls from the set of the first y balls?

Can you say how many?

To get $P(Y\le y)$, you need to divide this by the total number of ways you can draw n balls from the total set of r balls.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,793
Re: Probability-Probability!choose without replacement!

So is the answer:p(Y<=y)= {y choose n}/{r choose n}???
Yep!

- - - Updated - - -

And P(Z>=z)=1-P(Z<z)=1-{z choose n}/{r choose n} ???
Hold on. That's a bit tricky.
The result might be correct, but Z is different from Y.

Z is the smallest number, so we are talking about getting n balls from the set {z, z+1, ..., r}.
How many balls in that set?
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Re: Probability-Probability!choose without replacement!

So...is the answer P(Z>=z)={r-z+1 choose n}/{r choose n}???? (Wondering)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,793
Re: Probability-Probability!choose without replacement!

So...is the answer P(Z>=z)={r-z+1 choose n}/{r choose n}???? (Wondering)
Right! (Wink)
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Ok!!!Thank you!!! :cool: