A box has 5 white and 6 black balls. From the box we are taking out a sample without replacement and the sample size is chosen randomly from the set $\{2,3,4\}$. Calculate the probability for all the chosen balls being white.
I'm not quite sure how to approach this but I was thinking about first calculating the probabilities individually for each sample size and then multiplying them together.
$$P(A_k)=\frac{\dbinom Kk \dbinom {N-K}{n-k}}{\dbinom Nn} \qquad where \space k\in \Bbb N_0$$
$N=11 \qquad K=5 \qquad N-K=6$
where $K$ is the white balls and $N-K$ the black balls.
$A$ = all balls are white which means that $k=n$
$P(A_2)=\frac{\dbinom 52 \dbinom {6}{0}}{\dbinom {11}{2}}=3,006253 \cdot 10^{-6}$
$P(A_3)=\frac{\dbinom 53 \dbinom {6}{0}}{\dbinom {11}{3}}=3,00625 \cdot 10^{-6}$
$P(A_4)=\frac{\dbinom 54 \dbinom {6}{0}}{\dbinom {11}{4}}=3,006253 \cdot 10^{-6}$.
Now $P(A_2)\cdot P(A_3) \cdot P(A_4)=2,7116918 \cdot 10^{-17}$.
This solution doesn't seem right to me and I think I need to change my approach but I don't know how.
The individual probabilities were calculated wrongly: $$P(A_2)=\frac2{11}\qquad P(A_3)=\frac2{33}\qquad P(A_4)=\frac1{66}$$ You also need to take a weighted sum of the probabilities rather than multiplying them together. In this case, the sample size is one of $\{2,3,4\}$ with equal probability, so the correct probability is $$\frac{P(A_2)+P(A_3)+P(A_4)}3=\frac{\frac2{11}+\frac2{33}+\frac1{66}}3=\frac{17}{198}=0.0858585\dots$$