Answer
$P(A~royal~flush)=\frac{1}{649740}\approx0.000001539$
Work Step by Step
First, let's find $P(royal~flush~of~clubs)$
- First card:
The sample space are the 52 cards. So, $N(S_1)=52$
Let $E_1$ be the next event: "get one of the five following cards": Ten of clubs, Jack of clubs, Queen of clubs, King of clubs or Ace of clubs. $N(E_1)=5$
Using the Classical Method (page 259):
$P(E_1)=\frac{N(E_1)}{N(S_1)}=\frac{5}{52}$
- Second card:
The sample space are the 51 remaining cards. So, $N(S_2)=51$
Let $E_2$ be the next event: "get one of the four remaining of the five initial cards": $N(E_2~|~E_1)=4$
Using the Classical Method (page 259):
$P(E_2~|~E_1)=\frac{N(E_2~|~E_1)}{N(S_2)}=\frac{4}{51}$
- Third card:
The sample space are the 50 remaining cards. So, $N(S_3)=50$
Let $E_3$ be the next event: "get one of the three remaining of the five initial cards": $N(E_3~|~E_1~and~E_2)=3$
Using the Classical Method (page 259):
$P(E_3~|~E_1~and~E_2)=\frac{N(E_3~|~E_1~and~E_2)}{N(S_3)}=\frac{3}{50}$
- Fourth card:
The sample space are the 49 remaining cards. So, $N(S_4)=49$
Let $E_4$ be the next event: "get one of the two remaining of the five initial cards": $N(E_4~|~E_1~and~E_2~and~E_3)=2$
Using the Classical Method (page 259):
$P(E_4~|~E_1~and~E_2~and~E_3)=\frac{N(E_3~|~E_1~and~E_2~and~E_3)}{N(S_4)}=\frac{2}{49}$
- Fifth card:
The sample space are the 48 remaining cards. So, $N(S_5)=48$
Let $E_5$ be the next event: "get the last of the five initial cards": $N(E_5~|~E_1~and~E_2~and~E_3~and~E_4)=1$
Using the Classical Method (page 259):
$P(E_5~|~E_1~and~E_2~and~E_3~and~E_4)=\frac{N(E_5~|~E_1~and~E_2~and~E_3~and~E_4)}{N(S_5)}=\frac{1}{48}$
Now, using the General Multiplication Rule (page 289):
$P(royal~flush~of~clubs)=P(E_1)\times P(E_2~|~E_1)\times P(E_3~|~E_1~and~E_2)\times P(E_4~|~E_1~and~E_2~and~E_3)\times P(E_5~|~E_1~and~E_2~and~E_3~and~E_4)=\frac{5}{52}\times\frac{4}{51}\times\frac{3}{50}\times\frac{2}{49}\times\frac{1}{48}=\frac{120}{311875200}=\frac{1}{2598960}\approx0.0000003848$
The events "royal flush of clubs", "royal flush of diamonds", "royal flush of hearts" and "royal flush of spades" are mutually exclusive (disjoint events). Also:
$P(royal~flush~of~clubs)=P(royal~flush~of~diamonds)=P(royal~flush~of~hearts)=P(royal~flush~of~spades)=\frac{1}{2598960}$
Now, using the Addition Rule for Disjoint Events (page 270):
$P(A~royal~flush)=P(royal~flush~of~clubs)+P(royal~flush~of~diamonds)+P(royal~flush~of~hearts)+P(royal~flush~of~spades)=\frac{1}{2598960}+\frac{1}{2598960}+\frac{1}{2598960}+\frac{1}{2598960}=\frac{4}{2598960}=\frac{1}{649740}\approx0.000001539$
