Answer
$P(card~1~is~club~and~card~2~is~club)=\frac{1}{16}=0.0625$
Work Step by Step
If the sampling is done with replacement, the events are independent.
- The first card:
The sample space: 52 cards from a standard 52-card deck. So, $N(S_1)=52$
There are 13 club cards. So, $N(card~1~is~club)=13$
Using the Classical Method (page 259):
$P(card~1~is~club)=\frac{N(card~1~is~club)}{N(S_1)}=\frac{13}{52}=\frac{1}{4}$
- The second card:
The sample space: 52 cards from a standard 52-card deck. So, $N(S_2)=52$
There are 13 club cards. So, $N(card~2~is~club)=13$
Using the Classical Method (page 259):
$P(card~2~is~club)=\frac{N(card~2~is~club)}{N(S_2)}=\frac{13}{52}=\frac{1}{4}$
Now, using the Multiplication Rule for Independent Events (page 282):
$P(card~1~is~club~and~card~2~is~club)=P(card~1~is~club)\times P(card~2~is~club)=\frac{1}{4}\times\frac{1}{4}=\frac{1}{16}=0.0625$
