How Conditional Probability Works in a Sampling Experiment My textbook explains the conditional probability $$P(B|A)$$ is the probability that even B occurs in the reduced sample space A, as in the figure.

The book then gives the following example:
We randomly picks two television tubes from a shipment of 240 tubes of which 15 are defective. The probability that the first one is defective is $P(A)=15/240$ and the probability that the second one is defective given that the first one is defective is $P(B|A)=14/239$. Thus, the probability that both will be defective is $15/240 * 14/239$.
I don't understand how conditional probability works in that example, especially in regard to the above figure. More specifically,


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*What is the sample space in the experiment?

*What is the reduced sample space in the experiment?

 A: Let $T$ denote the $240$ tubes of the shipment. Then


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*$T = G \cup D$, where $G$ is the set of good ones and $D$ is the set of defective ones.

*It is given that $|D| = 15$, hence $|G| = 225$.

*The sample space is $S = \{(t_1, t_2)\; | t_1 \in T, t_2 \in T\setminus \{t_1\} \}$

*$A = \{(d, t)\; | d \in D, t \in T\setminus \{d\} \}$ - first one is defective

*$\Rightarrow P(A) = \frac{15\cdot 239}{240 \cdot 239} = \frac{15}{240}$

*$B = \{(t, d)\; | d \in D, t \in T\setminus \{d\} \}$ - second one is defective

*$A \cap B = \{(d_1, d_2)\; | d_1 \in D, d_2 \in D\setminus \{d_1\} \}$
Since you are dealing with $P(B|A)$, the reduced sample space is now $A$. This is so, because the conditional probability of $B$ given $A$ restricts the consideration only to events which are contained in $A$.
Now, in your example, they use the conditional probability of $A$ given $B$ to calculate the probability of $B \cap A$:
$$P(B|A) = \frac{P(B\cap A)}{P(A)} \Rightarrow P(B\cap A) = P(B|A)\cdot P(A) = \frac{14}{239}\cdot \frac{15}{240}$$
