1. ## 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?

2. ## Re: ball throwing question..

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.

3. ## Re: ball throwing question..

i forgot to mention that every hole can contain at most 6 balls

4. ## Re: ball throwing question..

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.