Red queens and black jacks are removed from a pack of 52 playing cards. Find the probability that the card drawn from the remaining cards is:
a card of clubs or an ace

Answer

To solve the problem of finding the probability that the card drawn from the remaining cards is a card of clubs or an ace, we will follow these steps:

### Step 1: Determine the total number of cards remaining
Initially, there are 52 cards in a standard deck. We have removed 2 red queens and 2 black jacks.

- Total cards removed = 2 (red queens) + 2 (black jacks) = 4 cards
- Remaining cards = 52 - 4 = 48 cards

### Step 2: Identify the number of clubs in the remaining cards
In a standard deck, there are 13 clubs. However, since we removed the black jacks, we need to account for that.

- Number of clubs removed = 1 (the black jack of clubs)
- Remaining clubs = 13 - 1 = 12 clubs

### Step 3: Identify the number of aces in the remaining cards
There are 4 aces in a standard deck, and none of the aces have been removed.

- Remaining aces = 4 aces

### Step 4: Determine the overlap between clubs and aces
One of the aces is a club (the ace of clubs). Therefore, we need to consider this overlap when calculating the probability.

- Number of clubs that are also aces = 1 (ace of clubs)

### Step 5: Use the formula for the probability of the union of two events
The probability of drawing a card that is either a club or an ace can be calculated using the formula:

\[ P(C \cup A) = P(C) + P(A) - P(C \cap A) \]

Where:
- \( P(C) \) = Probability of drawing a club
- \( P(A) \) = Probability of drawing an ace
- \( P(C \cap A) \) = Probability of drawing a card that is both a club and an ace

### Step 6: Calculate each probability
1. **Probability of drawing a club (P(C))**:
\[ P(C) = \frac{\text{Number of clubs}}{\text{Total remaining cards}} = \frac{12}{48} \]

2. **Probability of drawing an ace (P(A))**:
\[ P(A) = \frac{\text{Number of aces}}{\text{Total remaining cards}} = \frac{4}{48} \]

3. **Probability of drawing a card that is both a club and an ace (P(C ∩ A))**:
\[ P(C \cap A) = \frac{\text{Number of clubs that are also aces}}{\text{Total remaining cards}} = \frac{1}{48} \]

### Step 7: Substitute values into the formula
Now we can substitute these probabilities back into the union formula:

\[ P(C \cup A) = P(C) + P(A) - P(C \cap A) \]
\[ P(C \cup A) = \frac{12}{48} + \frac{4}{48} - \frac{1}{48} \]

### Step 8: Simplify the expression
Combine the fractions:

\[ P(C \cup A) = \frac{12 + 4 - 1}{48} = \frac{15}{48} \]

### Conclusion
The probability that the card drawn from the remaining cards is either a card of clubs or an ace is:

\[ \frac{15}{48} \]