- Thread starter
- #1

- Apr 13, 2013

- 3,836

Thanks in advance!!!!

- Thread starter evinda
- Start date

- Thread starter
- #1

- Apr 13, 2013

- 3,836

Thanks in advance!!!!

- Admin
- #2

- Mar 5, 2012

- 9,485

Hi evinda!

Thanks in advance!!!!

Your problem looks a bit odd.

Is there perhaps something missing from the problem statement?

Anyway, if you choose n balls, then you

So the probability of n balls is one, and the probability to anything other than n balls is zero.

- Thread starter
- #3

- Apr 13, 2013

- 3,836

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

- Admin
- #4

- Mar 5, 2012

- 9,485

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

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.

- Thread starter
- #5

- Apr 13, 2013

- 3,836

So is the answer(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.

- Admin
- #6

- Mar 5, 2012

- 9,485

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

- - - Updated - - -

Hold on. That's a bit tricky.And P(Z>=z)=1-P(Z<z)=1-{z choose n}/{r choose n} ???

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?

- Thread starter
- #7

- Apr 13, 2013

- 3,836

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

- Admin
- #8

- Mar 5, 2012

- 9,485

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

- Thread starter
- #9

- Apr 13, 2013

- 3,836

Ok!!!Thank you!!!