A bag contains 7 white balls and 3 red balls. A ball is taken and its colour noted. It is not replaced. A second ball... The question is: A bag contains 7 white balls and 3 red balls. A ball is taken and its colour noted. It is not replaced. A second ball is taken and its colour noted. Find the probability of obtaining two white balls.
Solution:
Let $A$ be the event 'the first ball is white'.
Let $B$ be the event 'the second ball is white'.
$P(A \cap B)=P(A) P(B \mid A)=\frac{7}{10} \times \frac{6}{9}=\frac{7}{15}$
I don't understand how $P(B \mid A)=\frac{6}{9}$. Intuitively, it makes sense: $\frac{6}{9}$ represents the probability that the second ball is white given that the first ball is white, and this is sampling without replacement.
But $P(B \mid A)=\frac{n(A \cap B)}{n(A)}$, so am I to understand that $n(A)=9$ and $n(A \cap B)=2$? The fact that $n(A)=9$ and $n(A \cap B)=2$ doesn't make sense to me, even though $P(B \mid A)=\frac{6}{9}$ seems intuitively correct. Thanks..

$n(A \cap B)$ and $n(A)$ are not counting the number of balls in the bag. They are counting the number of ways to satisfy the conditions of $A \cap B$ and $A$. That is:
$n(A \cap B)$ is the total number of all possible ways of getting both the first and second ball as whites. There are $7$ possible ways of getting the first white ball; then, after that white ball is removed, there are $6$ possible ways of getting the second white ball. So, $n(A \cap B) = 7 \cdot 6$.
$n(A)$ is the total number of all possible ways of getting the first ball as white. The second ball can be anything. So, there are $7$ ways of getting the first white ball; then, after that first white ball is removed from the bag, $10 - 1 = 9$ balls remain. So there are $9$ ways of getting the second ball. Hence $n(A) = 7 \cdot 9$.
Thus $n(A \cap B) / n(A) = (7 \cdot 6) / (7 \cdot 9) = 6/9$. And this brute force counting approach matches the more "intuitive" $P(B | A) = 6/9$.
