Question

If a card is drawn from a pack of cards, what is the probability that it is either a club or a jack?

• A. \(\frac{17}{52}\)
• B. \(\frac{1}{52}\)
• C. \(\frac{7}{13}\)
• D. \(\frac{4}{13}\)

Hint:

Add the probability of a club and a jack, then subtract the overlap, the jack of clubs, once.

Answer 1

The Correct Option is D

Step 1: Note the total number of cards.
A standard pack has 52 cards in total.

Step 2: Count the clubs and the jacks separately.
There are 13 clubs in the pack (one for each rank), and there are 4 jacks in total, one from each suit.

Step 3: Watch out for double counting.
The jack of clubs is counted in both groups, once as a club and once as a jack, so it must not be counted twice. Using the addition rule for two overlapping events:
\[ P(\text{club or jack}) = P(\text{club}) + P(\text{jack}) - P(\text{jack of clubs}) \]

Step 4: Substitute the individual probabilities.
\[ P(\text{club}) = \frac{13}{52}, \quad P(\text{jack}) = \frac{4}{52}, \quad P(\text{jack of clubs}) = \frac{1}{52} \]
\[ P(\text{club or jack}) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52} \]

Step 5: Simplify and compare with the options.
\[ \frac{16}{52} = \frac{4}{13} \]
Option (a) \(\frac{17}{52}\) is what you get if you forget to subtract the overlap, option (b) \(\frac{1}{52}\) is just the jack of clubs alone, and option (c) \(\frac{7}{13}\) does not match this calculation at all.

Final Answer:
The probability is \(\frac{4}{13}\).
\[ \boxed{\dfrac{4}{13}} \]