Question

A card is drawn at random from a standard deck of 52 cards. Then, the probability of getting either an ace or a club is:

• A. 17/52
• B. 16/52
• C. 1/4
• D. 1/12

Hint:

When dealing with "or" probabilities, always check if the events overlap. If they do (i.e., are not mutually exclusive), you must subtract the probability of the overlap to avoid double-counting.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem involves calculating the probability of the union of two events. Since the two events (drawing an ace and drawing a club) are not mutually exclusive (there is an ace of clubs), we must use the addition rule of probability.

Step 2: Key Formula or Approach:
The probability of event A or event B occurring is given by: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] where \(A \cap B\) represents the event that both A and B occur.

Step 3: Detailed Explanation:
Let's define the events:

Event A: The card drawn is an ace.
Event B: The card drawn is a club. \end{itemize} A standard deck has 52 cards.

There are 4 aces in the deck. So, the probability of drawing an ace is \(P(A) = \frac{4}{52}\).

There are 13 clubs in the deck. So, the probability of drawing a club is \(P(B) = \frac{13}{52}\).
\end{itemize} The events are not mutually exclusive because there is one card that is both an ace and a club: the Ace of Clubs.

The event \(A \cap B\) is drawing the Ace of Clubs. The probability of this is \(P(A \cap B) = \frac{1}{52}\).
\end{itemize} Now, we apply the addition rule: \[ P(\text{Ace or Club}) = P(A) + P(B) - P(A \cap B) \] \[ = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} \] \[ = \frac{4+13-1}{52} = \frac{16}{52} \]
Step 4: Final Answer:
The probability of getting either an ace or a club is \(\frac{16}{52}\).