Answer
$P(get~two~clubs~in~a~row)=\frac{7}{205}\approx0.03415$
Work Step by Step
- First card:
The sample space are the 41 cards that I do not know. So, $N(S_1)=41$
Now, consider the event "first card is a club". $N(first~card~is~a~club)=8$
Using the Classical Method (page 259):
$P(first~card~is~a~club)=\frac{N(first~card~is~a~club)}{N(S_1)}=\frac{8}{41}$
- Second card:
The sample space are the 40 remaining cards that I do not know. So, $N(S_2)=40$
Now, consider the event "second card is a club". $N(second~card~is~a~club~|~first~card~is~a~club)=7$
Using the Classical Method (page 259):
$P(second~card~is~a~club~|~first~card~is~a~club)=\frac{N(second~card~is~a~club~|~first~card~is~a~club)}{N(S_2)}=\frac{7}{40}$
Now, using the General Multiplication Rule (see page 289):
$P(get~two~clubs~in~a~row)=P(first~card~is~a~club)\times P(second~card~is~a~club~|~first~card~is~a~club)=\frac{8}{41}\times \frac{7}{40}=\frac{56}{1640}=\frac{7}{205}\approx0.03415$
