Two cards are drawn at random without replacement from a well shuffled standard deck of 52 cards. What is the probability that one of the cards is a club (any club) and the other card is a king of a non club suit?

Introduction / Context:
This question involves card probability with two different types of cards: any club and a king from a non club suit. The deck is standard with 52 cards, 13 in each suit and 4 kings in total. Because the cards are drawn without replacement, outcomes must be counted using combinations of two card hands rather than independent single card probabilities alone. We want exactly one club and one king of a non club suit, so the king of clubs is not allowed to count as both items at once.

Given Data / Assumptions:

Standard deck = 52 cards.

Clubs in the deck = 13 cards.

Kings in the deck = 4 cards.

King of clubs is excluded from the king choice, because we require a club card and a separate non club king.

Two cards are drawn at random without replacement.

The event of interest is: one card is a club (any rank) and the other is a king of hearts, diamonds or spades.

Concept / Approach:
We work with unordered pairs of cards because in a random two card hand, the order of drawing does not matter for the final combination. The total number of possible two card hands is 52C2. Favourable hands consist of one club and one non club king. Since the sets of clubs and non club kings are disjoint, we can count the favourable pairs simply by multiplying the number of possibilities for each card type.

Step-by-Step Solution:
Step 1: Total number of two card hands = 52C2.Step 2: Compute 52C2 = (52 * 51) / (2 * 1) = 1326.Step 3: Count the number of club cards: there are 13 clubs in the deck.Step 4: Count the number of kings that are not clubs. There are 4 kings in total and exactly 1 king of clubs, so non club kings = 4 - 1 = 3.Step 5: To form a favourable hand, choose 1 club and 1 non club king. Because the two groups are disjoint, number of favourable hands = 13 * 3 = 39.Step 6: Probability = favourable / total = 39 / 1326.Step 7: Simplify 39 / 1326 by dividing numerator and denominator by 3 to obtain 13 / 442.

Verification / Alternative check:
We could also compute the probability using an ordered approach and then adjust. For example, choose the club first and then the non club king, or vice versa. The probability for club then non club king is (13/52) * (3/51). The probability for non club king then club is (3/52) * (13/51). Adding these gives (13 * 3 / (52 * 51)) * 2 = 78 / 2652 = 39 / 1326 = 13 / 442, confirming the combination method.

Why Other Options Are Wrong:
The value 29/442 arises from counting an incorrect number of favourable pairs. The fractions 1/26 and 1/13 are much larger probabilities and correspond to more frequent events, such as drawing any king or drawing a club without restricting the second card. The value 3/52 is closer to the probability of drawing a king in one draw, but it does not incorporate the specific requirement of having both card types simultaneously. Only 13/442 is consistent with correct counting of two card hands.

Common Pitfalls:
One pitfall is to allow the king of clubs as part of the event, which would double count cases where a single card fulfils both conditions. Another is to mix up ordered and unordered reasoning, for example counting 39 favourable ordered outcomes but dividing by 52C2. Both ordered and unordered methods work, but the counting and denominator must match. Always carefully interpret the wording: here, the club and the king must be distinct cards, and the king must come from a non club suit.

Final Answer:
The probability that one card is a club and the other is a king of a non club suit is 13/442.