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:
Show 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.
Step 1: Concept Overview:
This problem calculates the probability of drawing an ace or a club from a standard deck of cards. Since these events aren't mutually exclusive, we use the addition rule of probability. Step 2: Formula:
The probability of event A or B is:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
where \(A \cap B\) is the probability of both A and B occurring. Step 3: Explanation:
Define the events:
Event A: Drawing an ace.
Event B: Drawing a club.
A standard deck contains 52 cards.
\(P(A) = \frac{4}{52}\) because there are 4 aces.
\(P(B) = \frac{13}{52}\) because there are 13 clubs.
The events aren't mutually exclusive due to the ace of clubs.
\(A \cap B\) is drawing the ace of clubs, so \(P(A \cap B) = \frac{1}{52}\).
Applying 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: Answer:
The probability of drawing an ace or a club is \(\frac{16}{52}\).