Q.A bag contains $5$ red and $4$ black balls. Two balls are drawn one after another without replacement. Find the probability that both are red.

"Without replacement" makes the second draw conditional on the first, so use the multiplication theorem.

There are $5+4=9$ balls. Let $R_1$ = "first ball is red" and $R_2$ = "second ball is red".

First draw: $P(R_1)=\dfrac{5}{9}$.

After removing one red (no replacement), $8$ balls remain of which $4$ are red, so the conditional probability

By the multiplication theorem,

$P(R_1\cap R_2)=P(R_1)\cdot P(R_2\mid R_1)=\frac{5}{9}\times\frac{4}{8}=\frac{20}{72}=\frac{5}{18}.$ …

The combination method $\dfrac{\binom{5}{2}}{\binom{9}{2}}$ ignores draw order and …

Using $\dfrac{5}{9}\times\dfrac{5}{9}$ — that is the with-replacement answer. Without replacement, the second fraction must use the reduced count …