How to distinguish conditional probablity $P(A|B)$ and $P(A\cap B)$ in some practical problems? A factory makes a type of computer monitor which are defective independently each with probability 1/20. To leave the factory, each monitor has to pass a test by a machine. The probability that the machine rejects a non-defective monitor is 1/10, and the probability that the machine passes a defective monitor is 1/20. What is the probablity that a typical monitor that passes the test is defective?

*

*I am confused about the differences between "the machine passes a
defective monitor" and "a typical monitor that passes the test is
defective". Which one is the conditional probability?

*Also, does the probablity of the machine passes a defective monitor denote P(pass the test and defective)?

 A: Update: Distinguishing Conditional Probability
The key indication when trying to identify whether a statement is a conditional probability or an unconditional probability statement is to see whether or not the statement restricts the universe of possibilities before asking the question.
For example, imagine a universe composed of only red balls and white balls. On each ball there is a positive integer. So we have two sets of probability states. The first is $R$ vs. $\bar{R}$ where the latter means not red but white. The second is $E$ vs. $\bar{E}$ where the latter means not even but odd.
Unconditional Statement
The following question is an unconditional probability statement "What is the probability of drawing an even-numbered white ball?". Note there is no restriction on the types of ball or number in the question. So the question is asking for $P(\bar{R}\cap E)$.
Conditional Statement
The following question is a conditional probability statement "What is the probability of of a white ball being even?". Note that even before we start, we have eliminated all the red balls from the universe. So here we are asking $P(E|\bar{R})$. This last is $\frac{P(\bar{R}\cap E)}{P(E)}$. The numerator is the same as in the unconditional statement. But as we are asking not for the probability of drawing an even white ball from all the balls, but the probability of a white ball being even, we've restricted ourselves to a subset of the total probability universe and now live in the conditional world.
Approaching this particular problem
Let's identify the probability states as follows:

*

*$G$: Machine is good

*$\bar{G}$: Machine is not good or defective

*$P$: Machine passes test

*$\bar{P}$: Machine does not pass test or fails

What you were given in your question is one unconditional and two conditional probability statements. The unconditional statement is:

*

*…which are defective independently each with probability $\frac{1}{20}$.

This means that having absolutely nothing to do with the test, there is a 5% chance that any one monitor will be defective. We write this as $P(\bar{G}) = 0.05$.
The first conditional statement is:

*

*The probability that the machine rejects a non-defective monitor is $\frac{1}{10}$
This means that given the machine is passed any number of of good monitors, it nevertheless will fail one out of every 10. As we are limiting ourselves to the universe of good monitors, this is a conditional statement and can be written as $P(\bar{P}|G) = 0.1$. This is also known as the False Negative Rate.
The next conditional statement can be parsed similarly:

*

*The probability that the machine passes a defective monitor is $\frac{1}{20}$.

This means that restricted to the universe of defective monitors, the probability of the test being incorrect and nevertheless passing the monitor is one out of twenty. This is a conditional statement which can be written as $P(P|\bar{G}) = 0.05$ and is also known as the False Positive Rate.
Now the question you are asked is:

*

*What is the probablity that a typical monitor that passes the test is defective?

This is saying, restricted to the universe of monitors which have passed the test rightly or wrongly, what is the probability that any one monitor is actually defective. Or in conditional probability terms: $P(\bar{G}|P)$.
To answer this, we usually use Bayes Theorem:
$$
P(A|B) = \frac{P(B|A)P(A)}{P(B)}
$$
Remembering that:
$$
P(A|B) = \frac{P(A\cap B)}{P(B)}
$$
So in our problem, we can write
$$
P(\bar{G}|P) = \frac{P(P|\bar{G})P(\bar{G})}{P(P)}
$$
We were given three pieces of information:

*

*$P(\bar{P}|G) = 0.1$: Conditional Statement 1

*$P(P|\bar{G}) = 0.05$:Conditional Statement 2

*$P(\bar{G}) = 0.05$: Unconditional Statement

We immediately know $P(G) = 0.95$ since the probability of an event AND its complement must be 100%. Now we have two out of the three pieces we need. All we need to find is $P(P)$. What's key to remember is the rule that since $G$ and $\bar{G}$ span the entire universe, $P(G\cap P) + P(\bar{G}\cap P) = P(P)$.
We can find $P(\bar{G}\cap P)$ as follows:
$$
\begin{aligned}
P(P|\bar{G}) &= 0.05\\
\frac{P(P\cap \bar{G})}{P(\bar{G})} &= 0.05\\
P(P\cap \bar{G}) &= 0.0025\\
\end{aligned}
$$
The last point to remember is that if $P(\bar{P}|G) = 0.1$ then $P(P|G) = 0.9$. Since once we restrict ourselves to the world of $G$, an event and its complement span the universe.
Using this, you can figure out the last missing piece, $P(P\cap G)$, find $P(P)$, and solve the Bayesian equation.
A: In your question a probability of any intersection is not present.

Let $D$ be the event of being defective and let $P$ denote the event of being passed.
Then the data are:

*

*$\Pr(D)=\frac1{20}$

*$\Pr(P^c|D^c)=\frac1{10}$

*$\Pr(P|D)=\frac1{20}$
Note that the second bullet is equivalent with:

*

*$\Pr(P|D^c)=\frac9{10}$
Based on this info you are asked to find $\Pr(D|P)$ and this can be done by applying the rule of Bayes.
Essential for this is the equality:
$$\Pr(D|P)\Pr(P)=\Pr(D\cap P)=\Pr(P|D)\Pr(D)$$
It tells you that finding $\Pr(P)$, $\Pr(D)$ and $\Pr(P|D)$ is enough for finding $\Pr(D|P)$.
Try this yourself.

In your first bullet you seem to wonder about $\Pr(P|D)$ and $\Pr(D|P)$. Both probabilities are conditional.
In your second bullet (if I understand that well) you seem to be wondering if we have $\Pr(P|D)=\Pr(P\cap D)$. No. LHS is the (conditional) probability that the test is passed if the monitor is defective and RHS is the probability that the test is passed and the monitor is defective.
I hope that the things I tried to make clear above have spread some light.
A: The answer to the question

How to distinguish conditional probability $(|)$ and $(\cap )$ in some practical problems?

is to read the problem carefully. Conditional probability is signaled by the last sentence

What is the probablity that a typical monitor that passes the test is
defective?

which asks for a probability (is it defective?) when you are given extra information (it passes the test).
The other probabilities are all for events that you can specify without an extra condition.
For a way work on problems like this with fewer formulas, see Applied Probability- Bayes theorem
