Two cards are drawn one by one at random from a deck of $52$ playing cards. What is the probability of getting two jacks if
- the first card is replaced before the second is drawn;
- the first card is not replaced before the second card is drawn.
Let $A$ be the event of drawing a jack in the first draw.
Let $B$ be the event of drawing a jack in the second draw.
- When the first card is replaced before the second card is drawn:
Number of elements in sample space $S = n \left(S\right) = 52$
$\because \;$ there are $4$ jacks
$n \left(A\right) = 4$, $n \left(B\right) = 4$
Event $A$ will not affect the probability of occurrence of event $B$
i.e. $\;$ $A$ and $B$ are independent events.
$\begin{aligned} \therefore \; \text{Required probability} = P \left(A \cap B\right) & = P \left(A\right) \times P \left(B\right) \\\\ & = \dfrac{n \left(A\right)}{n \left(S\right)} \times \dfrac{n \left(B\right)}{n \left(S\right)} \\\\ & = \dfrac{4}{52} \times \dfrac{4}{52} \\\\ & = \dfrac{1}{169} \end{aligned}$
- When the first card is not replaced before the second card is drawn:
In the first draw, there are $4$ jacks and $52$ cards in total.
$\because \;$ The card drawn is not replaced, for the second draw there are only $3$ jacks and $51$ cards in total.
$\therefore \;$ We have $P \left(A\right) = \dfrac{4}{52}$, $P \left(B | A\right) = \dfrac{3}{51}$
By definition, $P \left(B | A\right) = \dfrac{P \left(A \cap B\right)}{P \left(A\right)}$
$\begin{aligned} \therefore \; P \left(A \cap B\right) & = P \left(A\right) \times P \left(B | A\right) \\\\ & = \dfrac{4}{52} \times \dfrac{3}{51} \\\\ & = \dfrac{1}{221} \end{aligned}$