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

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The discussion focuses on calculating the probabilities P(Y<=y) and P(Z>=z) in a random selection of n balls from a total of r balls numbered 1 to r, chosen without replacement. The probability P(Y<=y) is determined by the formula P(Y<=y) = {y choose n}/{r choose n}, while P(Z>=z) is calculated as P(Z>=z) = {r-z+1 choose n}/{r choose n}. The participants confirm that these formulas accurately represent the probabilities based on the conditions of the selection process.

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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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