Problem source

In a certain factory, microchips are made by two machines.
Machine A makes a 
proportion $\lambda$ of the chips, where $0 < \lambda < 1$, and machine B makes the rest. 
A proportion $p$ of the chips  made by machine A are perfect, and
a  proportion $q$  of those made by machine B are perfect, 
where $0 < p < 1$ and $0 < q < 1$. The chips are sorted into two groups: group 1
contains those that are
perfect and group 2 contains those that are imperfect.
In a large random sample taken from group 1, it is found that
$\frac 2 5$ 
were made by machine A. Show that $\lambda$ can estimated as
\[
 {2q \over 3p + 2q}\;.
\]
Subsequently, it is discovered that the sorting process 
is faulty: there is a probability of $\frac 14$ that a perfect
 chip is assigned to group 2 and a  probability of $\frac 14$ that an imperfect
 chip is assigned to group 1. Taking into account this additional information,
obtain a new  estimate of $\lambda\,$.

Solution source

\begin{align*}
&& \frac25 &= \frac{\lambda p}{\lambda p + (1-\lambda) q} \\
\Rightarrow && 2(1-\lambda)q &= 3\lambda p \\
\Rightarrow && \lambda(3p+2q) &= 2q \\
\Rightarrow && \lambda &= \frac{2q}{3p+2q}
\end{align*}

\begin{align*}
&& \frac25 &= \frac{\lambda (p + \frac14(1-p))}{\lambda (p + \frac14(1-p))+(1-\lambda) (q + \frac14(1-q))} \\
&&&= \frac{\lambda(\frac34p + \frac14)}{\lambda(\frac34p + \frac14)+(1-\lambda)(\frac34q + \frac14)} \\
\Rightarrow && \lambda &= \frac{2(\frac34q+\frac14)}{3(\frac34p + \frac14)+2(\frac34q+\frac14)} \\
&&&= \frac{\frac32q + \frac12}{\frac94p + \frac32q + \frac54} \\
&&&= \frac{6q+2}{9p+6q+5}
\end{align*}