Answer
$P(all~3~cards~are~aces)=\frac{1}{5525}\approx0.0001810$
Work Step by Step
- First card:
The sample space are the 52 cards. So, $N(S_1)=52$
There are 4 aces. Now, consider the event "first card is an ace". $N(first~card~is~an~ace)=4$
Using the Classical Method (page 259):
$P(first~card~is~an~ace)=\frac{N(first~card~is~an~ace)}{N(S_1)}=\frac{4}{52}=\frac{1}{13}$
- Second card:
The sample space are the 51 remaining cards. So, $N(S_2)=51$
There are 3 remaining aces. Now, consider the event "second card is an ace". $N(second~card~is~an~ace~|~first~card~is~an~ace)=3$
Using the Classical Method (page 259):
$P(second~card~is~an~ace~|~first~card~is~an~ace)=\frac{N(second~card~is~an~ace~|~first~card~is~an~ace)}{N(S_2)}=\frac{3}{51}=\frac{1}{17}$
- Third card:
The sample space are the 50 remaining cards. So, $N(S_3)=50$
There are 2 remaining aces. Now, consider the event "third card is an ace". $N(third~card~is~an~ace~|~the~two~first~cards~are~aces)=2$
Using the Classical Method (page 259):
$P(third~card~is~an~ace~|~the~two~first~cards~are~aces)=\frac{N(third~card~is~an~ace~|~the~two~first~cards~are~aces)}{N(S_3)}=\frac{2}{50}=\frac{1}{25}$
Now, using the General Multiplication Rule (page 289):
$P(all~3~cards~are~aces)=P(first~card~is~an~ace)\times P(second~card~is~an~ace~|~first~card~is~an~ace)\times P(third~card~is~an~ace~|~the~two~first~cards~are~aces)=\frac{1}{13}\times\frac{1}{17}\times\frac{1}{25}=\frac{1}{5525}\approx0.0001810$
