Picking 4 balls with replacement.

Question

I'd like your help with the following exercise I found in a set of difficult -at least for me- exercises. There is a box that contains 10 White, 20 Black and 30 Red balls, we choose 4 with replacement.

a. What's the probability we chose at least one White ball?

b. What's the probability we chose at least one White ball taking into account we didn't choose any Red ones?(conditional)

c. What's the probability we didn't choose any Red balls, taking into account that we chose at least one White ball?(conditional)

So, b. and c. have to do with a. For a. I thought I could find the probability that I didn't choose any white balls.

The possible outcomes are: $\binom{20}4+\binom{20}3\binom{30}1+\binom{20}2\binom{30}2+\binom{20}1\binom{30}3+\binom{30}4$

Our sample space is: $\binom{60}4$

I don't know if I'm thinking of it correctly but that's what I've came up with so far.

Answer 1

You are choosing with replacement, so you should just think in terms of drawing with $\frac 16$ chance of getting white, $\frac 13$ black, $\frac 12$ red. The combinations assume you are drawing without replacement. You have $\frac 56$ chance not to get white on each draw, so the chance you don't get a white one is ???

Answer 2

I thought I could find the probability that I didn't choose any white balls.

Yes, that is the correct approach.   However, you are overcomplicating your count for the size of the favoured event.   (Also, the selection is with replacement.)

a. You wish to find probability for the complement of selecting 4 from 50 non-white balls when selecting any 4 from 60 balls with replacement.$$\mathsf P(W>0)~=~1-\left(50/60\right)^4$$ b. You wish to find the probability for the complement of selecting 4 from 20 black balls when selecting any 4 from 30 black-or-white balls.$$\mathsf P(W>0\mid R=0)~=~\underline{\qquad}$$ c. Use Bayes' Rule. $$\mathsf P(R=0\mid W>0) ~=~ \dfrac{\mathsf P(W>0\mid R=0)\,\mathsf P(R=0)}{\mathsf P(W>0)}$$