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

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 will analyze the situation and calculate the probability that all three drawn cards are queens after half of the cards have been removed.

### Step 1: Understand the problem
We start with a standard deck of 52 playing cards. Half of the cards (26 cards) are removed randomly. We need to find the probability that when we draw 3 cards from the remaining 26 cards, all of them are queens.

### Step 2: Define the events
Let’s define the events based on how many queens are removed:
- **Q0**: No queens are removed.
- **Q1**: One queen is removed.
- **Q2**: Two queens are removed.
- **Q3**: Three queens are removed.
- **Q4**: Four queens are removed.

### Step 3: Calculate the probabilities for each event
1. **For Q0 (0 queens removed)**:
- Remaining cards = 48 (all 4 queens are still there).
- The probability of drawing 3 queens from the remaining 48 cards:
\[ P(E|Q0) = \frac{\binom{4}{3}}{\binom{26}{3}} = \frac{4}{2600} = \frac{1}{650} \]

2. **For Q1 (1 queen removed)**:
- Remaining cards = 48 (3 queens left).
- The probability of drawing 3 queens from the remaining 48 cards:
\[ P(E|Q1) = 0 \quad \text{(impossible to draw 3 queens)} \]

3. **For Q2 (2 queens removed)**:
- Remaining cards = 48 (2 queens left).
- The probability of drawing 3 queens from the remaining 48 cards:
\[ P(E|Q2) = 0 \quad \text{(impossible to draw 3 queens)} \]

4. **For Q3 (3 queens removed)**:
- Remaining cards = 48 (1 queen left).
- The probability of drawing 3 queens from the remaining 48 cards:
\[ P(E|Q3) = 0 \quad \text{(impossible to draw 3 queens)} \]

5. **For Q4 (4 queens removed)**:
- Remaining cards = 48 (0 queens left).
- The probability of drawing 3 queens from the remaining 48 cards:
\[ P(E|Q4) = 0 \quad \text{(impossible to draw 3 queens)} \]

### Step 4: Calculate the overall probability
Now we need to calculate the total probability using the law of total probability:
\[ P(E) = P(E|Q0) \cdot P(Q0) + P(E|Q1) \cdot P(Q1) + P(E|Q2) \cdot P(Q2) + P(E|Q3) \cdot P(Q3) + P(E|Q4) \cdot P(Q4) \]

- **Calculating \(P(Q0)\)**:
- The number of ways to choose 26 cards from 52 is \(\binom{52}{26}\).
- The number of ways to choose 26 cards from the 48 non-queens is \(\binom{48}{26}\).
- Thus, \(P(Q0) = \frac{\binom{48}{26}}{\binom{52}{26}}\).

- **Calculating \(P(Q1)\)**:
- The number of ways to choose 1 queen and 25 non-queens:
\[ P(Q1) = \frac{\binom{4}{1} \cdot \binom{48}{25}}{\binom{52}{26}} \]

- **Calculating \(P(Q2)\)**, \(P(Q3)\), and \(P(Q4)\) will be similar and will yield 0 for \(P(E|Q2)\), \(P(E|Q3)\), and \(P(E|Q4)\).

### Step 5: Combine results
The only non-zero contribution to \(P(E)\) comes from \(Q0\):
\[ P(E) = \frac{4}{\binom{52}{26}} \cdot \frac{\binom{48}{26}}{\binom{26}{3}} \]

### Step 6: Final calculation
After simplifying the above expression, we find the final probability.