From a pack of 52 playing cards, half of the cards are randomly removed without looking at them. From the remaining cards, 3 cards are drawn randomly. The probability that all are king.

Options

A. `(1)/((25)(17)(13))`
B. `(1)/((25)(15)(13))`
C. `(1)/((52)(17)(13))`
D. `(1)/((13)(51)(17))`

Answer

To solve the problem step by step, we need to calculate the probability that all three drawn cards are kings after half of the cards from a standard deck of 52 playing cards have been removed.

### Step 1: Understand the situation
A standard deck has 52 cards, including 4 kings. When half of the cards are removed, 26 cards remain. We need to consider different scenarios based on how many kings were removed.

### Step 2: Define the cases
Let’s denote:
- \( E_0 \): No kings removed
- \( E_1 \): One king removed
- \( E_2 \): Two kings removed
- \( E_3 \): Three kings removed
- \( E_4 \): All four kings removed

### Step 3: Calculate probabilities for each case
1. **Case \( E_0 \)**: No kings removed.
- Remaining cards = 26, Kings = 4
- Probability of drawing 3 kings:
\[ P(F | E_0) = \frac{\binom{4}{3}}{\binom{26}{3}} = \frac{4}{2600} = \frac{1}{650} \]

2. **Case \( E_1 \)**: One king removed.
- Remaining cards = 26, Kings = 3
- Probability of drawing 3 kings:
\[ P(F | E_1) = \frac{\binom{3}{3}}{\binom{26}{3}} = \frac{1}{2600} \]

3. **Case \( E_2 \)**: Two kings removed.
- Remaining cards = 26, Kings = 2
- Probability of drawing 3 kings:
\[ P(F | E_2) = \frac{\binom{2}{3}}{\binom{26}{3}} = 0 \quad (\text{impossible}) \]

4. **Case \( E_3 \)**: Three kings removed.
- Remaining cards = 26, Kings = 1
- Probability of drawing 3 kings:
\[ P(F | E_3) = \frac{\binom{1}{3}}{\binom{26}{3}} = 0 \quad (\text{impossible}) \]

5. **Case \( E_4 \)**: All four kings removed.
- Remaining cards = 26, Kings = 0
- Probability of drawing 3 kings:
\[ P(F | E_4) = \frac{\binom{0}{3}}{\binom{26}{3}} = 0 \quad (\text{impossible}) \]

### Step 4: Calculate the probabilities of each case
The number of ways to choose 26 cards from 52 is:
\[ \binom{52}{26} \]
Now we need to find the probabilities of each case \( E_i \):
- \( P(E_0) = \frac{\binom{48}{26}}{\binom{52}{26}} \)
- \( P(E_1) = \frac{\binom{48}{25} \cdot \binom{4}{1}}{\binom{52}{26}} \)
- \( P(E_2) = \frac{\binom{48}{24} \cdot \binom{4}{2}}{\binom{52}{26}} \)
- \( P(E_3) = \frac{\binom{48}{23} \cdot \binom{4}{3}}{\binom{52}{26}} \)
- \( P(E_4) = \frac{\binom{48}{22} \cdot \binom{4}{4}}{\binom{52}{26}} \)

### Step 5: Combine the probabilities
Now, we can combine the probabilities:
\[ P(F) = P(F | E_0) P(E_0) + P(F | E_1) P(E_1) + P(F | E_2) P(E_2) + P(F | E_3) P(E_3) + P(F | E_4) P(E_4) \]
Since \( P(F | E_2) \), \( P(F | E_3) \), and \( P(F | E_4) \) are all 0, we only need to consider \( E_0 \) and \( E_1 \).

### Step 6: Final Calculation
Substituting the values:
\[ P(F) = \left(\frac{1}{650} \cdot \frac{\binom{48}{26}}{\binom{52}{26}}\right) + \left(\frac{1}{2600} \cdot \frac{\binom{48}{25} \cdot 4}{\binom{52}{26}}\right) \]

### Step 7: Simplify the expression
After simplification, we can find the final probability.