Probability of a 3 card hand (without replacement) Three cards are dealt one after the another from a well shuffled deck of 52 standard playing cards.
What is the probability of the event that:


*

*The first card is a Red card.

*The second card is either equal to seven or is a Spade

*The third card is either the King of Diamonds or is less than 2.


I was searching for hypergeometric expression, but it seems Card 2 intrudes on sample space of Card 1. Similary for Card 3 and 2. 
Hence unable to solve. 
Any help is appreciated.
 A: Consider the following events:
$ A_1: $ card 1 is a red 7.
$ A_2: $ card 1 is the King of Diamonds or is a red card less than 2.
$ A_3: $ card 1 is any red card not included in $ A_1 $ or $ A_2 $. 
$ B_1: $ card 2 is a spade less than 2.
$ B_2: $ card 2 is equal to seven or is a spade that is not les than 2.
$ C: $ card 3 is the King of Diamonds or is less than 2. 
We want to calculate the following probability: $ P((A_1 \cup A_2 \cup A_3) \cap (B_1 \cup B_2) \cap C) $.
Using the distributive property, this can be re-written as:
$$ P((A_1 \cap (B_1 \cup B_2) \cap C) \cup (A_2 \cap (B_1 \cup B_2) \cap C) \cup (A_3 \cap (B_1 \cup B_2) \cap C)) $$ 
$$ = P((A_1 \cap B_1 \cap C) \cup (A_1 \cap B_2 \cap C) \cup (A_2 \cap B_1 \cap C) \cup (A_2 \cap B_2 \cap C) \cup (A_3 \cap B_1 \cap C) \cup (A_3 \cap B_2 \cap C)) $$
$$ = P(A_1 \cap B_1 \cap C) + P(A_1 \cap B_2 \cap C) + P(A_2 \cap B_1 \cap C) + P(A_2 \cap B_2 \cap C) + P(A_3 \cap B_1 \cap C) + P(A_3 \cap B_2 \cap C) $$
Next, we can apply the multiplication rule to get:
$$ P(A_1) P(B_1|A_1) P(C|A_1 \cap B_1) + P(A_1) P(B_2|A_1) P(C|A_1 \cap B_2) + P(A_2) P(B_1|A_2) P(C|A_2 \cap B_1) + P(A_2) P(B_2|A_2) P(C|A_2 \cap B_2) + P(A_3) + P(B_1|A_3) P(C|A_3 \cap B_1) + P(B_2|A_3) P(C|A_3 \cap B_2) $$
Finally, we want to compute each of the probabilities in the above equation. It is tedious but doable, and you will arrive at the right answer... 
