To determine the probability that a card picked at random from a standard deck of playing cards is either a club or a jack, we can use the principle of set theory. Here's the step-by-step process:
1. Identify the Total Number of Cards in the Deck:
- A standard deck contains 52 cards.
2. Determine the Number of Clubs in the Deck:
- There are 13 clubs in a standard deck, one for each rank (2 through Ace).
3. Determine the Number of Jacks in the Deck:
- There are 4 jacks in a standard deck (Jack of Spades, Jack of Hearts, Jack of Diamonds, and Jack of Clubs).
4. Identify the Overlap Between Clubs and Jacks:
- One of the jacks is also a club, specifically the Jack of Clubs. This card has been counted both among the 13 clubs and the 4 jacks, so we must ensure we don't double-count it.
5. Calculate the Number of Cards that are Either Clubs or Jacks:
- Add the number of clubs and the number of jacks, then subtract the one card that is both a club and a jack to avoid double-counting:
Number of clubs = 13
Number of jacks = 4
Overlap (Jack of Clubs) = 1
So, the number of favorable outcomes (either a club or a jack) is:
[tex]\[
\text{Number of clubs} + \text{Number of jacks} - \text{Overlap} = 13 + 4 - 1 = 16
\][/tex]
6. Determine the Total Number of Possible Outcomes:
- There are 52 cards in total.
7. Calculate the Probability:
- The probability of drawing a card that is either a club or a jack is the number of favorable outcomes divided by the total number of possible outcomes:
[tex]\[
\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{16}{52}
\][/tex]
8. Simplify the Fraction:
- To simplify [tex]\(\frac{16}{52}\)[/tex], divide both the numerator and the denominator by their greatest common divisor, which is 4:
[tex]\[
\frac{16 \div 4}{52 \div 4} = \frac{4}{13}
\][/tex]
Therefore, the probability that a card picked at random from a standard deck of playing cards is a club or a jack is [tex]\(\frac{4}{13}\)[/tex].
So, the correct answer is:
A. [tex]\(\frac{4}{13}\)[/tex]
