(related to Problem: Broken Items in the Box II)
We can apply the urn model without replacement, in which for the event
\[A:=\{\text{"If we draw }2\text{ items in total, then we draw exactly }k\text{ broken items."}\}\]
the probability obeys the equation
\[p_k=p(A)=\frac{\binom 4k\binom {10-4}{2-k}}{\binom {10}2}=\frac{\binom 4k\binom {6}{2-k}}{45}\quad\quad\text{for }k=0,1,2.\]
It follows \[p_0=\frac{\binom 40\binom {6}{2-0}}{45}=\frac{1\cdot 15}{45}=\frac{5}{15}=\frac{1}{3},\] \[p_1=\frac{\binom 41\binom {6}{2-1}}{45}=\frac{4\cdot 6}{45}=\frac{8}{15},\] \[p_2=\frac{\binom 42\binom {6}{2-2}}{45}=\frac{6\cdot 1}{45}=\frac{2}{15}.\]
