(related to Problem: Broken Items in the Box)
We define the events
\(A:="\text{The first item drawn is not broken}"\)
\(B:="\text{The second item drawn is not broken}"\)
Because we have a Laplace experiment, the probabilities of \(A\) and \(\overline{A}\) are
\[p(A)=\frac {10-4}{10}=\frac 35;\quad\quad p(\overline{A})=1-p(A)=\frac 25.\]
Because the drawing is without replacement, the probabilities of \(B\) given \(A\) and \(\overline{A}\) are
\[p(B|A)=\frac {9-4}{9}=\frac 59;\quad\quad p(B|\overline{A})=\frac {9-3}{9}=\frac 69=\frac 23.\]
Note that the events \(A\) and \(\overline{A}\) are mutually exclusive and collectively exhaustive. It follows that we can apply the law of total probability and calculate the probability of \(B\):
\[p(B)=p(B|A)p(A)+ p(B|\overline{A})p(\overline{A})=\frac 56\cdot \frac 35+\frac 23\cdot \frac 25=\frac{27}{45}=\frac 35.\]
It turns out that the events \(A\) and \(B\) have the same probability.