Answer
$P(Five~clubs)=\frac{33}{66640}\approx0.0004952$
Work Step by Step
- First card:
The sample space are the 52 cards. So, $N(S_1)=52$
There are 13 club cards. Now, consider the event "First card is a club". $N(First~card~is~a~club)=13$
Using the Classical Method (page 259):
$P(First~card~is~a~club)=\frac{N(First~card~is~a~club)}{N(S_1)}=\frac{13}{52}=\frac{1}{4}$
- Second card:
The sample space are the 51 remaing cards. So, $N(S_2)=51$
There are 12 remaining club cards. Now, consider the event "Second card is a club". $N(Second~card~is~a~club~|~First~card~is~a~club)=12$
$P(Second~card~is~a~club~|~First~card~is~a~club)=\frac{N(Second~card~is~a~club~|~First~card~is~a~club)}{N(S_2)}=\frac{12}{51}=\frac{4}{17}$
- Third card:
The sample space are the 50 remaing cards. So, $N(S_3)=50$
There are 11 remaining club cards. Now, consider the event "Third card is a club". $N(Third~card~is~a~club~|~Two~first~cards~are~club)=11$
$P(Third~card~is~a~club~|~Two~first~cards~are~club)=\frac{N(Third~card~is~a~club~|~Two~first~cards~are~club)}{N(S_3)}=\frac{11}{50}$
- Fourth card:
The sample space are the 49 remaing cards. So, $N(S_4)=49$
There are 10 remaining club cards. Now, consider the event "Fourth card is a club". $N(Fourth~card~is~a~club~|~Three~first~cards~are~club)=10$
$P(Fourth~card~is~a~club~|~Three~first~cards~are~club)=\frac{N(Fourth~card~is~a~club~|~Three~first~cards~are~club)}{N(S_4)}=\frac{10}{49}$
- Fifth card:
The sample space are the 48 remaing cards. So, $N(S_5)=48$
There are 9 remaining club cards. Now, consider the event "Fifth card is a club". $N(Fifth~card~is~a~club~|~Four~first~cards~are~club)=9$
$P(Fifth~card~is~a~club~|~Four~first~cards~are~club)=\frac{N(Fifth~card~is~a~club~|~Four~first~cards~are~club)}{N(S_5)}=\frac{9}{48}=\frac{3}{16}$
Now, using the General Multiplication Rule (page 289):
$P(Five~clubs)=P(First~card~is~a~club)\times P(Second~card~is~a~club~|~First~card~is~a~club)\times P(Third~card~is~a~club~|~Two~first~cards~are~club)\times P(Fourth~card~is~a~club~|~Three~first~cards~are~club)\times P(Fifth~card~is~a~club~|~Four~first~cards~are~club)=\frac{1}{4}\times\frac{4}{17}\times\frac{11}{50}\times\frac{10}{49}\times\frac{3}{16}=\frac{1320}{2665600}=\frac{33}{66640}\approx0.0004952$
