Solving a permutations-combinations problem

#1
Hi
Could you please explain to me how to do the following question?

"There are 12 beads in an urn, where 5 of them are blue, 4 are red and 3 are yellow. Beads with the same colour are identical. You pick out 9 beads randomly from the urn without replacement.

a) Find the probability that there are exactly 2 colours of beads chosen.

What I know: This means that you have to pick out the 5 blue and 4 red beads, meaning that each time, you can't pick out the yellow. This means that there is a probability of (9/12)*(8/11)*(7/10)... giving 9!4!/12!

b) Find the probability that there are exactly 3 blue beads chosen

This is where I get stuck. I simply have no idea what I am supposed to do here!

c) Suppose now that you randomly pick out 9 beads from the urn, but with replacement. Find the probability of observing exactly two colours of beads, red and yellow, in the 9 chosen beads. (Hint: Find the probability that no beads chosen are blue first.)

I am not entirely sure what to do here. The probability of not selecting a blue is 7/12. Do I just do (7/12)^9 as each time I am making the same selection?

Many thanks
 
Last edited:

Dason

Ambassador to the humans
#2
Hi! :welcome: We are glad that you posted here! This looks like a homework question though. Our homework help policy can be found here. We mainly just want to see what you have tried so far and that you have put some effort into the problem. I would also suggest checking out this thread for some guidelines on smart posting behavior that can help you get answers that are better much more quickly.
 
#3
Hi
Could you please explain to me how to do the following question?

"There are 12 beads in an urn, where 5 of them are blue, 4 are red and 3 are yellow. Beads with the same colour are identical. You pick out 9 beads randomly from the urn without replacement.

a) Find the probability that there are exactly 2 colours of beads chosen.

What I know: This means that you have to pick out the 5 blue and 4 red beads, meaning that each time, you can't pick out the yellow. This means that there is a probability of (9/12)*(8/11)*(7/10)... giving 9!4!/12!

b) Find the probability that there are exactly 3 blue beads chosen

This is where I get stuck. I simply have no idea what I am supposed to do here!

c) Suppose now that you randomly pick out 9 beads from the urn, but with replacement. Find the probability of observing exactly two colours of beads, red and yellow, in the 9 chosen beads. (Hint: Find the probability that no beads chosen are blue first.)

I am not entirely sure what to do here. The probability of not selecting a blue is 7/12. Do I just do (7/12)^9 as each time I am making the same selection?

Many thanks
Thank you for your response. I have updated the post with what I know and look forward to a response!
 

BGM

TS Contributor
#4
1. Apply the (multi)hypergeometric pmf. Note that the question has not restrict to red and yellow only, so you have to consider the other combinations. Moreover, you need to exclude the 1-color case.

2. Apply the hypergeometric pmf. The red and yellow beans can be merged as a non-blue group.

3. Yes by independence you can calculate the probability of no blue in that way. Again you need to be cautious to remove the 1-color case.