Two cards are drawn one by without replacement from a well shuffled deck of 52 cards. Find the proability distribution of the number of (I) Aces (II) kings (III) face cards (IV) spades.

Answer

To solve the problem of finding the probability distribution of the number of Aces, Kings, Face Cards, and Spades when drawing two cards from a standard deck of 52 cards without replacement, we can follow these steps:

### Step 1: Understand the Total Cards and Their Types
A standard deck has:
- 4 Aces
- 4 Kings
- 12 Face Cards (4 Kings, 4 Queens, 4 Jacks)
- 13 Spades

### Step 2: Define the Random Variable
Let \( X \) be the random variable representing the number of Aces, Kings, Face Cards, or Spades drawn in two draws.

### Step 3: Calculate Probability Distribution for Aces

#### Case 1: \( P(X = 0) \) (No Aces)
- Total ways to choose 2 cards from 48 non-Ace cards:
\[ \text{Ways} = \binom{48}{2} = \frac{48 \times 47}{2} = 1128 \]
- Total ways to choose 2 cards from 52:
\[ \text{Total ways} = \binom{52}{2} = \frac{52 \times 51}{2} = 1326 \]
- Probability:
\[ P(X = 0) = \frac{1128}{1326} = \frac{188}{221} \]

#### Case 2: \( P(X = 1) \) (1 Ace)
- Choose 1 Ace from 4 and 1 non-Ace from 48:
\[ \text{Ways} = \binom{4}{1} \times \binom{48}{1} = 4 \times 48 = 192 \]
- Probability:
\[ P(X = 1) = \frac{192}{1326} = \frac{32}{221} \]

#### Case 3: \( P(X = 2) \) (2 Aces)
- Choose 2 Aces from 4:
\[ \text{Ways} = \binom{4}{2} = 6 \]
- Probability:
\[ P(X = 2) = \frac{6}{1326} = \frac{1}{221} \]

### Summary for Aces
\[ \begin{array}{|c|c|} \hline X & P(X) \\ \hline 0 & \frac{188}{221} \\ 1 & \frac{32}{221} \\ 2 & \frac{1}{221} \\ \hline \end{array} \]

### Step 4: Calculate Probability Distribution for Kings
The calculations for Kings will follow the same structure as for Aces since there are also 4 Kings.

#### Case 1: \( P(X = 0) \)
\[ P(X = 0) = \frac{188}{221} \]

#### Case 2: \( P(X = 1) \)
\[ P(X = 1) = \frac{32}{221} \]

#### Case 3: \( P(X = 2) \)
\[ P(X = 2) = \frac{1}{221} \]

### Summary for Kings
\[ \begin{array}{|c|c|} \hline X & P(X) \\ \hline 0 & \frac{188}{221} \\ 1 & \frac{32}{221} \\ 2 & \frac{1}{221} \\ \hline \end{array} \]

### Step 5: Calculate Probability Distribution for Face Cards
There are 12 Face Cards.

#### Case 1: \( P(X = 0) \)
- Choose 2 from 40 non-Face Cards:
\[ P(X = 0) = \frac{\binom{40}{2}}{\binom{52}{2}} = \frac{780}{1326} = \frac{130}{221} \]

#### Case 2: \( P(X = 1) \)
- Choose 1 Face Card from 12 and 1 non-Face Card from 40:
\[ P(X = 1) = \frac{12 \times 40}{1326} = \frac{480}{1326} = \frac{80}{221} \]

#### Case 3: \( P(X = 2) \)
- Choose 2 Face Cards from 12:
\[ P(X = 2) = \frac{\binom{12}{2}}{\binom{52}{2}} = \frac{66}{1326} = \frac{11}{221} \]

### Summary for Face Cards
\[ \begin{array}{|c|c|} \hline X & P(X) \\ \hline 0 & \frac{130}{221} \\ 1 & \frac{80}{221} \\ 2 & \frac{11}{221} \\ \hline \end{array} \]

### Step 6: Calculate Probability Distribution for Spades
There are 13 Spades.

#### Case 1: \( P(X = 0) \)
- Choose 2 from 39 non-Spade Cards:
\[ P(X = 0) = \frac{\binom{39}{2}}{\binom{52}{2}} = \frac{741}{1326} = \frac{247}{442} \]

#### Case 2: \( P(X = 1) \)
- Choose 1 Spade from 13 and 1 non-Spade from 39:
\[ P(X = 1) = \frac{13 \times 39}{1326} = \frac{507}{1326} = \frac{169}{442} \]

#### Case 3: \( P(X = 2) \)
- Choose 2 Spades from 13:
\[ P(X = 2) = \frac{\binom{13}{2}}{\binom{52}{2}} = \frac{78}{1326} = \frac{26}{442} \]

### Summary for Spades
\[ \begin{array}{|c|c|} \hline X & P(X) \\ \hline 0 & \frac{247}{442} \\ 1 & \frac{169}{442} \\ 2 & \frac{26}{442} \\ \hline \end{array} \]