Probability of Dealing cards

Question

Randomly distributing 10 cards from a standard deck of 52 cards to 2 players, Xavier and Zorro, so each player gets 5 cards. (a) If Xavier has two Ace cards, find the probability that Zorro has at least one Ace card. (b) Is the event that Xavier gets an Ace independent of the event that Zorro gets a Spade.

For part (a), If X already had 2 Aces, there are only 2 more Aces left for Z to choose from. For Z to have at least 1 Ace, Z can have 1 out of 2 Aces and 4 out of 48 non-Ace cards or both Aces and 3 non-Ace cards. So the number of possibilities is C(2,1)C(48,4) + C(2,2)C(48,3) = 406,456. However, the number of possibilities that X has 2 Aces is C(2,2)C(48,3) = 17,296. This is how far I can do. I've learned the conditional prob that says P(A|B) = P(A and B)/P(B) but not quite sure what is the difference between P(Z|X) and P(Z and X) for this problem.

For part (b), I know that to proof independence, I need to proof that P(A|B) = P(A). P(X gets an Ace) = C(4,1)C(48,4)/C(52,5) = 778,320/2,598,960 = 0.2995. As for P(X gets an Ace | Z gets a Spade), I'm not sure how to do because there is one card which is an Ace of Spade. But my sense is they may not be independent, just can't proof it mathematically.

Answer 1

For part (a), If X already had 2 Aces, there are only 2 more Aces left for Z to choose from. For Z to have at least 1 Ace, Z can have 1 out of 2 Aces and 4 out of 48 non-Ace cards or both Aces and 3 non-Ace cards. So the number of possibilities is C(2,1)C(48,4) + C(2,2)C(48,3) = 406,456. However, the number of possibilities that X has 2 Aces is C(2,2)C(48,3) = 17,296. This is how far I can do.

You do not need the last.

You have counted favoured outcomes (Zorro is dealt at least one ace; when Xavier has exactly two ace and three non-ace).

You merely need to count total outcomes (Zorro is dealt any five from forty-seven cards; when Xavier has two aces and three non ace).

Then divide, as normal (after being assure that all outcomes are equally probable).

$$\dfrac{\mathrm C(2,1)\,\mathrm C(45,4) + \mathrm C(2,2)\,\mathrm C(45,3)}{\mathrm C(47,5)}$$

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Note: you could also have used that the event of Zorro receiving no aces is the complement of the favoured event.

$$\dfrac{\mathrm C(47,5)-\mathrm C(45,5)}{\mathrm C(47,3)}$$

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Alternatively we can count ways to place the aces among either the second hand or the remaining forty-two cards in the deck.$$\dfrac{\mathrm C(5,1)\,\mathrm C(42,1)+\mathrm C(5,2)}{\mathrm C(47,2)}$$

These are, of course, equal.

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For part (b), I know that to proof independence, I need to proof that P(A|B) = P(A). P(X gets an Ace) = C(4,1)C(48,4)/C(52,5) = 778,320/2,598,960 = 0.2995. As for P(X gets an Ace | Z gets a Spade), I'm not sure how to do because there is one card which is an Ace of Spade. But my sense is they may not be independent, just can't proof it mathematically.

Use the law of total probability, partitioning on whether the Ace of Spades is in Xavier's hand or not.

$${\mathsf P(\spadesuit\in Z\mid A\in X)}={{\mathsf P(\spadesuit\in Z\mid A\spadesuit\in X)\mathsf P(A\spadesuit\in X\mid A\in X)}\\+{\mathsf P(\spadesuit\in Z\mid A\in X, A\spadesuit\notin X)\mathsf P(A\spadesuit\notin X\mid A\in X)}}$$

And of course $\mathsf P(A\spadesuit\in X)=\dfrac{\mathsf P(A\spadesuit\in X)}{\mathsf P(A\in X)}$ and such...