Probability of a 3 card hand (without replacement)

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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:

  1. The first card is a Red card.
  2. The second card is either equal to seven or is a Spade
  3. 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.

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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...