# ball throwing question..

• Dec 22nd 2011, 05:24 AM
transgalactic
ball throwing question..
i have 6 balls .
balls 1-3 are white
balls 4,5 are red
ball 6 is blue

we throw these balls into 5 holes numerated 1-5
what is the probability that in hole number 3 we get only red balls
and there is no red balls in the other holes

i saw another student soluton and it says
$\frac{1^24^4}{5^6}$

cant understand the numerator
we need to choose 1 red ball out of 2 for hole number 3,so its $\binom{2}{1}$

divided by $5^6$

am i correct?
• Dec 22nd 2011, 06:26 AM
Plato
Re: ball throwing question..
Quote:

Originally Posted by transgalactic
i have 6 balls .
balls 1-3 are white, balls 4,5 are red, ball 6 is blue
we throw these balls into 5 holes numerated 1-5
what is the probability that in hole number 3 we get only red balls
and there is no red balls in the other holes
i saw another student soluton and it says
$\frac{1^24^4}{5^6}$ cant understand the numerator

That answer assumes a function model.
There are $5^6$ functions from a set of six to a set of five.
There is one way to put the balls 4&5 into hole #3.
There are $4^4$ ways to put the other balls into the other holes.
• Dec 22nd 2011, 06:39 AM
transgalactic
Re: ball throwing question..
i forgot to mention that every hole can contain at most 6 balls
• Dec 22nd 2011, 07:17 AM
Plato
Re: ball throwing question..
Quote:

Originally Posted by transgalactic
i forgot to mention that every hole can contain at most 6 balls

That fact does not change the answer in any way.
All six balls can be in the same hole.