If 2 cards are drawn from a pack of 52 playing cards, probability that at least one of them is ace is

Options

A. `1/5`
B. `3/16`
C. `9/20`
D. `33/221`

Answer

To find the probability that at least one of the two cards drawn from a pack of 52 playing cards is an Ace, we can use the complementary approach. Instead of calculating the probability of getting at least one Ace directly, we will first calculate the probability of not getting any Aces and then subtract that from 1.

### Step-by-Step Solution:

1. **Total Number of Cards**:
There are 52 cards in total in a standard deck.

2. **Number of Aces**:
There are 4 Aces in a deck of cards.

3. **Number of Non-Ace Cards**:
The number of non-Ace cards is \(52 - 4 = 48\).

4. **Calculate the Probability of Not Drawing an Ace**:
- When the first card is drawn, the probability that it is not an Ace is \(\frac{48}{52}\).
- When the second card is drawn (assuming the first card drawn was not an Ace), there are now 51 cards left, of which 48 are still non-Aces. Hence, the probability that the second card is also not an Ace is \(\frac{47}{51}\).

5. **Calculate the Combined Probability of Not Drawing Aces**:
The probability of both cards not being Aces is:
\[ P(\text{no Aces}) = \frac{48}{52} \times \frac{47}{51} \]

6. **Calculate the Probability of At Least One Ace**:
To find the probability of getting at least one Ace, we subtract the probability of not getting any Aces from 1:
\[ P(\text{at least one Ace}) = 1 - P(\text{no Aces}) = 1 - \left(\frac{48}{52} \times \frac{47}{51}\right) \]

7. **Simplifying the Calculation**:
- First, calculate \(P(\text{no Aces})\):
\[ P(\text{no Aces}) = \frac{48 \times 47}{52 \times 51} = \frac{2256}{2652} \]
- Now, simplify \(\frac{2256}{2652}\):
\[ \frac{2256 \div 12}{2652 \div 12} = \frac{188}{221} \]
- Therefore,
\[ P(\text{at least one Ace}) = 1 - \frac{188}{221} = \frac{221 - 188}{221} = \frac{33}{221} \]

### Final Answer:
The probability that at least one of the two cards drawn is an Ace is \(\frac{33}{221}\).