Probabilities of obtaining $k\in\{0,1,2,3,4\}$ white balls Two baskets containing $n_1$, respectively $n_2$ balls, of which $k_1 \lt n_1$ , respectively $k_2 \lt n_2$ are white. From the first basket a ball is extracted and is put in the other basket, from which there are extracted three balls consecutively(returned). Calculate the probabilities of obtaining $k\in\{ 0,1,2,3,4\}$ white balls.
I can't get my head around this one. Thanks!
 A: If k is the sum of white balls coming from basket one and/or basket two after this experiment than you use "laplace" for each basket:
$$P(k_{1})=\frac{number\; of \;cases\; white\; balls\; are\; extracted}{number\; of \;cases\; arbitrary\; ball\; is\; extraced\;}\; =\frac{k_{1}}{n_{1}}$$
for basket two the index is 2 but basically you use the same assumption.
No.1 $$P(k=0)$$
The probability to extract 0 white balls from basket one is
$$P(k_{1}=0)=\frac{number\; of \;cases\; white\; balls\; are\; extracted}{number\; of \;cases\; arbitrary\; ball\; is\; extraced\;}\; =1-\frac{k_{1}}{n_{1}} $$
after this you extract a ball (three times) from basket two knowing that there is not a white ball from basket one:
first time you extract from basket two:$$P(k_{2}=0)=\frac{number\; of \;cases\; white\; balls\; are\; extracted}{number\; of \;cases\; arbitrary\; ball\; is\; extraced\;}\; =1-\frac{k_{2}}{n_{2}+1} $$
another thing you have to "know" is that the first time you extract is independent from the second or third time you extract from basket two! These Three Probabilities are multiplied because of this!
so you end up for  $$P(k_{}=0)=P(k_{1}=0)+P(k_{2}=0)*P(k_{2}=0)*P(k_{2}=0)=$$
$$1-\frac{k_{1}}{n_{1}}+\left (1-\frac{k_{2}}{n_{2}+1}  \right )^{3}$$
for $$P(k_{}=1,2,3)$$ you just have to use the same ideas splitting the experiment:
note that for the other cases you basically have multiple possibilities which basket the white ball(s) can come from. (sum them up)
A: The wording of the problem is not entirely clear. We give one interpretation, and show how to calculate under that interpretation. 
We assume that the second drawing is done with replacement. Because of the presence of the possibility $k=4$, we assume that we are interested in the total number of white balls seen. That includes the white ball from the first urn, if a white ball was drawn.
Let $X$ be the total number of white balls seen. As a sample, we calculate $\Pr(X=2)$.
There are two possibilities for the first ball (i) white, with probability $\frac{k_1}{n_1}$ and (ii) not white, with probability $\frac{n_1-k_1}{n_1}$.
Given the first ball drawn is white, the probability the total of white seen is $2$ is the probability we get exactly $1$ white on the second draw. This is 
$\binom{3}{1}\left(\frac{k_2+1}{n_2+1}\right)^1 \left(1-\frac{k_2+1}{n_2+1}\right)^2$.
Given the first ball drawn is not white, the probability the total of white seen is $2$ is the probability we get exactly $2$ white on the second draw. This is 
$\binom{3}{2}\left(\frac{k_2}{n_2+1}\right)^2 \left(1-\frac{k_2}{n_2+1}\right)^1$.
Thus the probability that $X=2$ is
$$\frac{k_1}{n_1}\binom{3}{1}\left(\frac{k_2+1}{n_2+1}\right)^1 \left(1-\frac{k_1+1}{n_1}\right)^2+\frac{n_1-k_1}{n_1}\binom{3}{2}\left(\frac{k_2}{n_2+1}\right)^2 \left(1-\frac{k_2}{n_2+1}\right)^1.$$
We can find similar expressions for $\Pr(X=k)$ for $k=0,1,3,4$. 
