There is no exact closed form solution. If $k$ is moderately large, the normal approximation to the binomial will give you an approximate answer, and you can then do a numerical search around there for the exact answer. For the normal approximation we want $$\Phi\left(\frac{k-ns}{\sqrt{ns(1-s)}}\right) \geq .9,$$ where $\Phi$ is the normal cdf. You can take the inverse cdf, simplify this to a quadratic in $\sqrt n$, and find a value for $n$. Since this is not exact, you can then check other nearby values of $n$ to find the exact solution.
Here's an R function that solves it for you using this approach, though not particularly efficiently:
binomial.size <- function(k,s,p) {
# Find the smallest sample size n such that a binomial(n,s) has
# probability at least p of having k sucesses.
# our first approximation comes from solving the quadratic:
n <- ceiling(((-qnorm(p)+sqrt(qnorm(p)^2+4*k/(1-s)))/(2*sqrt(s/(1-s))))^2)
while(pbinom(k-1,n,s,lower.tail=FALSE)<p) { # our n might be too small...
n <- n+1
}
while(pbinom(k-1,n-1,s,lower.tail=FALSE)>p) { # or too large...
n <- n-1
}
return(n)
}
binomial.size(15,0.4,0.5)
## 37
pbinom(14,37,0.4,lower.tail=FALSE)
## 0.54
pbinom(14,36,0.4,lower.tail=FALSE)
## 0.48--too small!
Your answer for (i) is correct. Think about it combinatorially. We have $\binom{6}{4}$ ways to choose our green balls, and $\binom{10}{4}$ ways to choose $4$ balls. So we divide out:
$$\frac{ \binom{6}{4} } { \binom{10}{4} }$$
For (ii), we want one of each color. So we have one red ball- there are $\binom{4}{1}$ ways to do this. Then there are $\binom{6}{1}$ ways to choose a green ball. Finally, we have $8$ balls left so we choose $2$ balls: $\binom{8}{2}$. We divide out by the number of ways to choose $4$ balls.
(iii) I would look at this in terms of complementation. We have one red and one green ball. What are the odds of getting two red balls (ie., no more green balls)? Subtract this from $1$.
Edit: So we have one of each color has been drawn. So we have three red balls and 5 green balls left. What are the odds of drawing two red balls? We have:
$$\frac{ \binom{3}{2} }{\binom{8}{2} }$$
We choose $2$ red balls and no green balls, then divide out by the number of ways to draw two more balls.
So this is the complement of your solution. So you should easily be able to get the desired solution by subtracting this from $1$.
Best Answer
The count of favoured items in a sample selected from a population without replacement has a hypergeometric distribution.
When the population is size $N$ with $K$ favoured items, and the sample is of size $n$, then the count $W$ of favoured items in the sample having size $k$ has probability:
$$\mathsf P(W=k) = \dfrac{\dbinom{K}{k}~\dbinom{N-K}{n-k}}{\dbinom{N}{n}} \qquad \Big[0\leq k\leq \min(K, n) \leq \max(K, n) \leq N\Big]$$
$$\mathsf P(W\geqslant k) = \sum_{x=k}^{\min(K, n)} \dfrac{\dbinom{K}{x}~\dbinom{N-K}{n-x}}{\dbinom{N}{n}} \qquad \Big[0\leq k\leq \min(K, n) \leq \max(K, n) \leq N\Big]$$
For particular values it might be more appropriate to use an approximation, or work with the complement, to ease the computation load.
This is easiest calculated using the complement. It is the probability of not drawing zero white balls.
$$\mathsf P(W\geqslant 1) = 1-\mathsf P(W=0) = 1-\dfrac{\binom{250}{0}\binom{249750}{8500}}{\binom{250000}{8500}}\\ \approx 0.999{\small 825266071400062267418017708833099206480271885565627713\ldots}$$