Q.A bag contains $4$ red and $6$ black balls. Two balls are drawn one after the other without replacement. Find the probability that the first ball drawn is red and the second is black.

[!TLDR] $P(R_1)=\tfrac4{10}$, then $P(B_2\mid R_1)=\tfrac69$ (one red ball removed). [!ANSWER] $P(R_1\cap B_2)=\dfrac{4}{15}$.

$P(R_1)=\tfrac{4}{10}=\tfrac25$. After removing one red ball, $9$ balls remain with $6$ black, so $P(B_2\mid R_1)=\tfrac69=\tfrac23$. By the multiplication theorem, $P(R_1\cap B_2)=\tfrac25\times\tfrac23=\tfrac{4}{15}$. [!ANSWER] $P(R_1\cap B_2)=\dfrac{4}{15}$.

Multiply $P(\text{first red})$ by $P(\text{second black}\mid\text{first red})$, adjusting the total ball count after the first draw.

Using $10$ balls (instead of the remaining $9$) as the denominator for the second draw ignores that the pool shrinks after the first ball is removed.