There are 5 machines in a factory, out of which 3 are working correctly. They produce 10% of the defective items independently. Remaining 2 machines have 20% defective item production.
A machine is chosen at random and five articles produced by the machine are examined. What is the probability that the machine chosen is defective given that of the 5 articles examined 2 are imperfect and 3 are perfect.
In the solution, at the beginning they have calculated
$P(D\mid M) = {^5C_2} \cdot (0.1)^2\cdot (0.9)^3$.
Here $D$ stands for picking 3 perfect and 2 imperfect articles and $M$ stands for machine working correctly.
Here I don't understand why
$P(D \cap M) = {^5C_2} \cdot (0.1)^2\cdot (0.9)^3$. is incorrect?
If it's incorrect, what is the difference between $P(D\mid M)$ and $P(D \cap M)$ in this question?
I am getting confused when to use conditional probability and intersection in such cases.
$P(D|M)$ is the probability that a perfect machine produces two faulty items out of five.
$P(D\cap M)$ is the probability that a perfect machine is picked and it's produced two faulty items out of five.
Those two probabilities are linked via Bayes' formula: $P(D\cap M)=P(D|M)P(M)$, and so the latter is $\frac{3}{5}$ (which is $P(M)$) of the former, i.e.
$P(D\cap M)=\frac{3}{5}{5\choose 2}(0.1)^2(0.9)^3$