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: List the favourable cards directly instead of using the addition rule.
We want cards that are a club, a jack, or both. Rather than adding and subtracting probabilities, we can count the distinct cards that satisfy this directly.

Step 2: Count all 13 clubs first.
All 13 cards of the club suit qualify, since each one is a club: spades, hearts, diamonds and clubs each have 13 cards, and we take the full club group. This gives 13 cards so far.

Step 3: Add the jacks that are not already clubs.
There are 4 jacks in total: jack of clubs, jack of spades, jack of hearts, jack of diamonds. The jack of clubs is already included in the 13 clubs counted above, so we only add the remaining 3 jacks (spades, hearts, diamonds) that were not counted yet.

Step 4: Add up the distinct favourable cards.
\[ 13 \text{ clubs} + 3 \text{ new jacks} = 16 \text{ favourable cards} \]

Step 5: Divide by the total number of cards.
\[ P(\text{club or jack}) = \frac{16}{52} = \frac{4}{13} \]
Counting the distinct cards this way naturally avoids double counting the jack of clubs, since it is placed in only one of the two groups, and gives the same result as the formula-based approach.
\[ \boxed{\dfrac{4}{13}} \]