Question

If a card is drawn at random from a well shuffled pack of playing cards, then the probability that it is either an Ace or a Spade card is

• A. $\frac{4}{13}$
• B. $\frac{1}{13}$
• C. $\frac{1}{52}$
• D. $\frac{17}{52}$

Hint:

Remember the principle of inclusion-exclusion for "or" probabilities: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.

Answer 1

The Correct Option is A

Step 1: Identify the total number of possible outcomes.
A standard pack of playing cards contains 52 cards. So, the total number of possible outcomes is 52.
Step 2: Identify the number of favorable outcomes for the card being an Ace.
There are 4 Aces in a deck of cards.
Step 3: Identify the number of favorable outcomes for the card being a Spade.
There are 13 Spades in a deck of cards.
Step 4: Identify the number of favorable outcomes for the card being both an Ace and a Spade.
There is 1 card that is both an Ace and a Spade (the Ace of spades).
Step 5: Calculate the number of favorable outcomes for the card being either an Ace or a Spade.
Using the principle of inclusion-exclusion:
$$ \text{Number of (Ace or Spade)} = \text{Number of (Ace)} + \text{Number of (Spade)} - \text{Number of (Ace and Spade)} $$ $$ \text{Number of (Ace or Spade)} = 4 + 13 - 1 = 16 $$
Step 6: Calculate the probability of the card being either an Ace or a Spade.
$$ P(\text{Ace or Spade}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{16}{52} $$
Step 7: Simplify the probability.
$$ P(\text{Ace or Spade}) = \frac{16 \div 4}{52 \div 4} = \frac{4}{13} $$
Thus, the probability that the card drawn is either an Ace or a Spade card is $ \boxed{\frac{4}{13}} $.