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 calculate the probability that all three cards drawn from the remaining cards are queens after half of the cards are randomly removed.

### Step 1: Understand the Problem
We start with a standard deck of 52 playing cards, which includes 4 queens. When half of the cards (26 cards) are removed, we need to determine the probability that all three cards drawn from the remaining 26 cards are queens.

### Step 2: Define Events
Let’s define some events based on the number of queens that could have been removed:
- **E1**: No queens removed.
- **E2**: One queen removed.
- **E3**: Two queens removed.
- **E4**: Three queens removed.
- **E5**: All four queens removed.

### Step 3: Calculate Probabilities for Each Event
We will calculate the probability of drawing 3 queens given each event.

1. **Event E1 (No queens removed)**:
- Probability of selecting 3 queens from the remaining 4 queens:
\[ P(F|E1) = \frac{\binom{4}{3}}{\binom{26}{3}} = \frac{4}{2600} = \frac{1}{650} \]

2. **Event E2 (One queen removed)**:
- Probability of selecting 3 queens from the remaining 3 queens:
\[ P(F|E2) = \frac{\binom{3}{3}}{\binom{26}{3}} = \frac{1}{2600} \]

3. **Event E3 (Two queens removed)**:
- Probability of selecting 3 queens from the remaining 2 queens:
\[ P(F|E3) = 0 \quad \text{(impossible to select 3 from 2)} \]

4. **Event E4 (Three queens removed)**:
- Probability of selecting 3 queens from the remaining 1 queen:
\[ P(F|E4) = 0 \quad \text{(impossible to select 3 from 1)} \]

5. **Event E5 (All four queens removed)**:
- Probability of selecting 3 queens from none:
\[ P(F|E5) = 0 \quad \text{(impossible to select 3 from 0)} \]

### Step 4: Calculate Overall Probability
Now we need to use the law of total probability to find the overall probability of drawing 3 queens:
\[ P(F) = P(E1)P(F|E1) + P(E2)P(F|E2) + P(E3)P(F|E3) + P(E4)P(F|E4) + P(E5)P(F|E5) \]

### Step 5: Calculate Probabilities of Events E1, E2, E3, E4, E5
To find \(P(E1)\), \(P(E2)\), etc., we need to consider how many queens could be in the removed cards.

- The total ways to choose 26 cards from 52 is \(\binom{52}{26}\).
- The number of ways to choose 26 cards such that \(k\) queens are removed is given by:
\[ P(E_k) = \frac{\binom{4}{k} \cdot \binom{48}{26-k}}{\binom{52}{26}} \]

### Step 6: Combine Everything
Now we can substitute the values of \(P(E1)\), \(P(E2)\), etc., into the equation for \(P(F)\) and compute the final probability.

### Step 7: Final Calculation
After performing the calculations, we find that:
\[ P(F) = \frac{1}{13} \cdot \frac{1}{17} \cdot \frac{1}{25} \]

### Conclusion
The final probability that all three drawn cards are queens is:
\[ \frac{1}{650} \]