What Are the Probabilities P(Y<=y) and P(Z>=z) in a Random Ball Selection?

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Discussion Overview

The discussion revolves around calculating the probabilities P(Y<=y) and P(Z>=z) in a scenario involving the random selection of balls from a container. Participants explore the implications of choosing n balls without replacement and the definitions of Y as the greatest number and Z as the smallest number among the selected balls.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions whether there is something missing from the problem statement, suggesting that the probability of selecting n balls is always one.
  • Another participant clarifies that to find P(Y<=y), one must calculate the combinations of drawing n balls from the first y balls and divide this by the total combinations from r balls.
  • There is a proposal that P(Y<=y) could be expressed as {y choose n}/{r choose n}.
  • Discussion arises regarding P(Z>=z), with a participant suggesting it could be calculated as 1 - P(Z
  • A later reply proposes that P(Z>=z) might be {r-z+1 choose n}/{r choose n}, which receives affirmation from another participant.

Areas of Agreement / Disagreement

Participants express differing views on the calculations for P(Y<=y) and P(Z>=z), with some proposing specific formulas while others question the assumptions underlying these calculations. The discussion remains unresolved regarding the exact formulations and interpretations of the probabilities.

Contextual Notes

Participants do not fully agree on the assumptions needed for the calculations, particularly regarding the definitions of Y and Z and how they affect the probabilities. There are also unresolved questions about the implications of selecting balls without replacement.

evinda
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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!
 
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Re: Probability-Probability!choose without replacement!

evinda said:
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.
 
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:
 
Re: Probability-Probability!choose without replacement!

evinda said:
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. :o

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.
 
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} ?
I like Serena said:
Oh wait! I misread. :o

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.
 
Re: Probability-Probability!choose without replacement!

evinda said:
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?
 
Re: Probability-Probability!choose without replacement!

So...is the answer P(Z>=z)={r-z+1 choose n}/{r choose n}? (Wondering)
 
Re: Probability-Probability!choose without replacement!

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

Right! (Wink)
 
Ok!Thank you! :cool:
 

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