Probability of 3 aces from drawing 7 cards. 1.58 Intro to Probability, 2nd Ed.

Question

This is Problem 1.58 from Tsitsiklis, Bertsekas, Introduction to Probability, 2nd edition.

We draw the top 7 cards from a well-shuffled standard 52-card deck.
Find the probability that:
a) The 7 cards include exactly 3 aces.
b) The 7 cards include exactly 2 kings.
c) The probability that the 7 cards include exactly 3 aces, or exactly 2 kings, or both.

My solution was

a) $${3! {49 \choose 4} \over {52 \choose 7} }$$

b) $${2! {50 \choose 5} \over {52 \choose 7} }$$

c) $${3! {49 \choose 4} \over {52 \choose 7} } + {2! {50 \choose 5} \over {52 \choose 7} } - {3! 2! {47 \choose 2} \over {52 \choose 7} }$$

The solution book says

Why do they ${4 \choose 3}$ instead of (3!)?

We have 7 slots where we can put the cards with each slot accepting 1 card. We want to fill the first 3 with aces and the last 4 with non-aces. There are 3! ways to distribute aces into the first 3 slots.

And then there are ${49 \choose 4}$ ways to fill the remaining slots.

There is another card problem where they use that same logic to solve the problem:

Answer 1

a) Divide all cards into two groups: a group of $4$ aces and a group of $48$ non-ace cards. From the first group you select any $3$ aces and from the second group you select any $4$ cards.

b) Again divide all cards into two groups: a group of $4$ kings and a group of $48$ non-king cards. From the first group you select any $2$ kings and from the second group you select any $5$ cards.

c) Use inclusion-exclusion principle. Adding a) and b) will double count the outcomes of exactly 3 aces and 2 kings, which must be subtracted once. To find the outcomes of exactly 3 aces and 2 kings, you divide all cards into $3$ groups: a group of $4$ aces, a group of $4$ kings and a group of $44$ non-ace and non-king cards. From the first group you select any $3$ aces, from the second you select any $2$ kings and from the last group you select any $2$ cards.

Answer 2

You're not filling "the first three slots" with the aces, you're just deciding which three aces are among the first seven cards, and which four or the remaining $48$ non-aces fill out the first seven. Note that it's $48$, not $49$ as in your solution, since you can't allow the fourth ace.

Answer 3

(a)

We know that the total number of ways in which we can pick and arrange 7 out of 52 cards is given by: ${}^{52} \mathrm{ P }_7.$

Then, for picking 3 out of 4 aces in a standard deck of cards, there are ${}^4 \mathrm{ C }_3$ ways. Similarly, for picking 4 out of 48 cards (52 - 4 aces), there are ${}^{48} \mathrm{ C }_4$ ways. Finally, there are 7! ways to arrange the 7 cards that we have chosen.

So, the equation becomes:

$$=\frac{{}^4 \mathrm{ C }_3\times{}^{48} \mathrm{ C }_4\times7!}{{}^{52} \mathrm{ P }_7}$$

$$=\frac{{}^4 \mathrm{ C }_3\times{}^{48} \mathrm{ C }_4\times7!}{{}^{52} \mathrm{ C}_7\times7!}$$

$$=\frac{{}^4 \mathrm{ C }_3\times{}^{48} \mathrm{ C }_4}{{}^{52} \mathrm{ C}_7}.$$

(b)

Similar to the previous question,

We know that the total number of ways in which we can pick and arrange 7 out of 52 cards are given by: ${}^{52} \mathrm{ P }_7$ .

Then, for picking 2 out of 4 kings in a standard deck of cards, there are ${}^4 \mathrm{ C }_2$ ways. Similarly, for picking 5 out of 48 cards (52 - 4 kings), there are ${}^{48} \mathrm{ C }_5$ ways. Finally, there are 7! ways to arrange the 7 cards that we have chosen.

So, the equation becomes:

$$=\frac{{}^4 \mathrm{ C }_2\times{}^{48} \mathrm{ C }_5\times7!}{{}^{52} \mathrm{ P }_7}$$

=$\frac{{}^4 \mathrm{ C }_2\times{}^{48} \mathrm{ C }_5\times7!}{{}^{52} \mathrm{ C}_7\times7!}$

=$\frac{{}^4 \mathrm{ C }_2\times{}^{48} \mathrm{ C }_5}{{}^{52} \mathrm{ C}_7}$

(c)

In this part, we are required to find $P(A \cup B)$, where A is the event of choosing 7 cards with exactly 3 aces, and B is the event of choosing 7 cards with exactly 2 kings.

We know,

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

In the earlier sections, we computed P(A) and P(B). Now, we need to compute P(A$\cap$B). P(A$\cap$B) requires that there are exactly 2 kings and exactly 3 aces from the seven cards that we pick.

The total number of ways in which we can pick and arrange 7 out of 52 cards are given by: ${}^{52} \mathrm{ P }_7$

There are ${}^4 \mathrm{ C }_2$ ways to pick exactly two out of the four kings in a standard deck, ${}^4 \mathrm{ C }_3$ ways to pick 3 out of 4 aces, and ${}^{44} \mathrm{ C }_2$ ways to pick 2 out of the 44 remaining cards that are not kings or aces (44 = 52 cards - 4 aces - 4 kings). Finally, there are 7! ways to arrange the 7 cards that we have chosen.

Thus,

P(A$\cap$B) = $\frac{{}^4 \mathrm{ C }_2\times{}^4 \mathrm{ C }_3\times{}^{44} \mathrm{ C }_2\times7!}{{}^{52} \mathrm{ P }_7}$

P(A$\cap$B) = $\frac{{}^4 \mathrm{ C }_2\times{}^4 \mathrm{ C }_3\times{}^{44} \mathrm{ C }_2\times7!}{{}^{52} \mathrm{ C }_7\times7!}$

P(A$\cap$B) = $\frac{{}^4 \mathrm{ C }_2\times{}^4 \mathrm{ C }_3\times{}^{44} \mathrm{ C }_2}{{}^{52} \mathrm{ C }_7}$

Therefore, the final equation becomes:

$\frac{{}^4 \mathrm{ C }_3\times{}^{48} \mathrm{ C }_4}{{}^{52} \mathrm{ C}_7}$ + $\frac{{}^4 \mathrm{ C }_2\times{}^{48} \mathrm{ C }_5}{{}^{52} \mathrm{ C}_7}$ - $\frac{{}^4 \mathrm{ C }_2\times{}^4 \mathrm{ C }_3\times{}^{44} \mathrm{ C }_2}{{}^{52} \mathrm{ C }_7}$

$$= \frac{({}^4 \mathrm{ C }_3\times{}^{48} \mathrm{ C }_4) + ({}^4 \mathrm{ C }_2\times{}^{48} \mathrm{ C }_5) - ({}^4 \mathrm{ C }_2\times{}^4 \mathrm{ C }_3\times{}^{44} \mathrm{ C }_2)}{{}^{52} \mathrm{ C }_7}.$$

Hope this helps!