I have three problems about sampling without replacement, where the next is generalization of the other.
We are picking $k$ balls without replacement from an ur with $n_1$ white balls and $n_2$ black balls (so that the total amount of balls are $n = n_1 + n_2$ and $k < n$). Each white ball is weighted with a probability $p_1$ to be picked, and each black ball is weighted with a probability $p_2 = 1-p_1$ to be picked.
1) Assume that there is only $n_1 =1$ white ball and the rest are black, that is, $n_2 = n-1 > 0$. When picking $k$ balls (without replacement), what is the probability that atleast one of these are the white one?
2) (Generalization of 1) Assume that there are $n_1 \geq 1$ white balls and the rest are black, that is, $n_2 = n-n_1 > 0$. When picking $k$ balls (without replacement), what is the probability that atleast one of these are the white one?
3) (Generalization of 2) Assume that there are $n_1$ white balls, but there might be zero white balls ($n_1$ might equal $0$) and the rest are black ($n_2 = n-n_1 > 0$). When picking $k$ balls (without replacement), what is the probability that atleast one of these are the white one? Assume that the probability of white balls being in the urn is $p_0$.
I guess 1) can be calculated using complements and 3) be calculated with conditional probabilties, assuming that you know 2), but I'm not sure.
I appreciate all help I can get.
EDIT: My suggested solutions:
EDIT2: Clarified my question and fixed my suggested solution:
1) and 2)
\begin{align*} P(\text{atleast one white}) &= 1 - P(\text{no white}) \\ &= \frac{n_2p_2}{n}\cdot\frac{(n_2-1)p_2}{n-1}\cdot\ldots\cdot\frac{(n_2-k+1)p_2}{n-k+1} \\ &= 1 - \prod^k_{i=1}\frac{(n_2-i+1)p_2}{n-i+1} \end{align*}
3) \begin{align*} P(\text{atleast one white}) &= 1 - P(\text{no white}) \\ &= 1 - \left(P(n_1>0)P(\text{no white}| n_1>0)+P(n_1=0)P(\text{no white}| n_1=0)\right)\\ &= 1 - \left(p_0\cdot\left(\prod^k_{i=1}\frac{(n_2-i+1)p_2}{n-i+1}\right) + (1-p_0)\cdot1\right)\\ &= p_0 - p_0\cdot\left(\prod^k_{i=1}\frac{(n_2-i+1)p_2}{n-i+1}\right) \end{align*}
Could anyone confirm that this is correct?