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.

  • $\begingroup$ These are separate problems or one single event? $\endgroup$ – lulu Jan 4 '17 at 11:21
  • $\begingroup$ Assuming it's a single event, I'd break it into cases. Case I : Red, $7 \clubsuit$,Black Ace. And so on. Tedious, but there aren't actually all that many cases and they are mostly exclusive. $\endgroup$ – lulu Jan 4 '17 at 11:24
  • $\begingroup$ Please provide the following information: How many red cards? How many 7s? How many Spades? How many Kings? How many Diamonds? What values are considered less than $2$? $\endgroup$ – barak manos Jan 4 '17 at 11:33
  • $\begingroup$ @ Lulu: This is a single event. $\endgroup$ – pratchit Jan 6 '17 at 11:42
  • $\begingroup$ @ barak manos: There are 26 red cards, four 7s, 13 Spades, four Kings and 13 Diamonds. Ace is the only value that is less than 2. $\endgroup$ – pratchit Jan 6 '17 at 11:43

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...

  • 1
    $\begingroup$ Shouldn't the case where card 2 is a spade have an impact on the probability of event C? That is, card two could be a card less than two (the ace of spades) and still meet the condition of event 2 in the question. $\endgroup$ – DBPriGuy Jan 6 '17 at 23:40
  • $\begingroup$ @DBPriGuy Yes, you're right. Thanks for pointing that out! I've edited my post. $\endgroup$ – Michael R Jan 7 '17 at 2:12
  • $\begingroup$ I think the approach is correct. After a while I understood inclusion exclusion principle only will help. Will try to calculate and get back to u guys with the answer. Thanks a lot. $\endgroup$ – pratchit Jan 13 '17 at 5:21

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