Answer
$P(card~1~is~club~and~card~2~is~club)=\frac{4}{68}=0.05882$
Work Step by Step
- 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: the 51 remaining cards from a standard 52-card deck. So, $N(S_2)=51$
There are the 12 remaining club cards. So, $N(card~2~is~club~|~card~1~is~club)=12$
Using the Classical Method (page 259):
$P(card~2~is~club~|~card~1~is~club)=\frac{N(card~2~is~club~|~card~1~is~club)}{N(S_2)}=\frac{12}{51}=\frac{4}{17}$
Now, using the General Multiplication Rule (page 289):
$P(card~1~is~club~and~card~2~is~club)=P(card~1~is~club)\times P(card~2~is~club~|~card~1~is~club)=\frac{1}{4}\times\frac{4}{17}=\frac{4}{68}=0.05882$
