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

[evinda]

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!

 

[I like Serena]

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]

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)?

 

[I like Serena]

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)?

Oh wait! I misread.

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]

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.

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.

 

[I like Serena]

Re: Probability-Probability!choose without replacement!

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

Yep!

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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]

Re: Probability-Probability!choose without replacement!

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

 

[I like Serena]

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]

Ok!Thank you!