A card is drawn from a well shuffled pack of playing cards. What is the probability that it is either a spade or an ace or both?

Answer

To find the probability that a card drawn from a well-shuffled pack of playing cards is either a spade or an ace or both, we can follow these steps:

### Step 1: Define the Events
Let:
- \( A \) be the event that the card drawn is a spade.
- \( B \) be the event that the card drawn is an ace.

### Step 2: Calculate the Probability of Event A
In a standard deck of 52 playing cards, there are 13 spades. Therefore, the probability of drawing a spade is:
\[ P(A) = \frac{\text{Number of Spades}}{\text{Total Number of Cards}} = \frac{13}{52} \]

### Step 3: Calculate the Probability of Event B
There are 4 aces in a standard deck of cards (one for each suit: hearts, diamonds, clubs, and spades). Thus, the probability of drawing an ace is:
\[ P(B) = \frac{\text{Number of Aces}}{\text{Total Number of Cards}} = \frac{4}{52} \]

### Step 4: Calculate the Probability of the Intersection of Events A and B
The intersection of events \( A \) and \( B \) (i.e., drawing a card that is both a spade and an ace) is the ace of spades. There is only 1 ace of spades in the deck. Therefore:
\[ P(A \cap B) = \frac{\text{Number of Ace of Spades}}{\text{Total Number of Cards}} = \frac{1}{52} \]

### Step 5: Apply the Formula for the Union of Two Events
To find the probability that a card drawn is either a spade or an ace or both, we use the formula for the union of two events:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Substituting the values we calculated:
\[ P(A \cup B) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} \]

### Step 6: Simplify the Expression
Now, we simplify the expression:
\[ P(A \cup B) = \frac{13 + 4 - 1}{52} = \frac{16}{52} \]
This can be further simplified:
\[ P(A \cup B) = \frac{4}{13} \]

### Final Answer
The probability that the card drawn is either a spade or an ace or both is:
\[ \frac{4}{13} \]

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