Q.A bag contains $5$ red and $3$ black balls. Two balls are drawn one after another without replacement. Find the probability that both balls drawn are red.

[!TLDR] $P(R_1)=\tfrac58$, $P(R_2\mid R_1)=\tfrac47$ (one red ball removed). [!ANSWER] $P(\text{both red})=\dfrac{5}{14}$.

There are $5$ red and $3$ black balls, $8$ in all. $P(R_1)=\tfrac58$. After one red ball is removed without replacement, $7$ balls remain with $4$ red, so $P(R_2\mid R_1)=\tfrac47$. By the multiplication theorem, $P(R_1\cap R_2)=P(R_1)P(R_2\mid R_1)=\tfrac58\times\tfrac47=\tfrac{20}{56}=\tfrac{5}{14}$. [!ANSWER] $P(\text{both red})=\dfrac{5}{14}$.

Multiply the probability of the first draw by the conditional probability of the second draw given the first, updating the pool size and red-ball count after the first draw.

Treating the two draws as independent (using $P(R_1)\times P(R_1)=(5/8)^2$) ignores that the pool changes after the first ball is removed — this is only valid for draws with replacement.