Two cards are drawn at random from a well shuffled pack of 52 playing cards. What is the probability that either both cards are red or both cards are queens?

Introduction / Context:
This question combines two events for card selection. We draw two cards and want the probability that the pair is either both red or both queens. Because two queens can also both be red, we must handle the overlap between these events carefully using the union rule in probability.

Given Data / Assumptions:

• A standard deck has 52 cards.
• There are 26 red cards, consisting of hearts and diamonds.
• There are 4 queens in total.
• Among these queens, 2 are red queens and 2 are black queens.
• Two cards are drawn together without replacement, and all unordered pairs are equally likely.
• Event A: both cards are red.
• Event B: both cards are queens.

Concept / Approach:
We use combinations to count outcomes. The total number of possible pairs is C(52, 2). To find the probability of A or B, we use P(A or B) = P(A) + P(B) minus P(A and B). For counting P(A and B), we note that both cards must be red queens, so both must be chosen from the two red queens available.

Step-by-Step Solution:
Total pairs = C(52, 2) = 52 * 51 / 2 = 1326. Number of red cards = 26. Pairs with both red cards = C(26, 2) = 26 * 25 / 2 = 325. Number of queens = 4, so pairs with both queens = C(4, 2) = 4 * 3 / 2 = 6. Red queens are 2 in number, so pairs that are both red and both queens = C(2, 2) = 1. Favourable pairs = 325 + 6 - 1 = 330. Required probability = 330 / 1326. Simplify 330 / 1326 by dividing numerator and denominator by 6 to get 55 / 221.

Verification / Alternative check:
We can also compute probabilities directly. P(both red) = (26/52) * (25/51). P(both queens) = (4/52) * (3/51). P(both red queens) = (2/52) * (1/51). Then P(A or B) = P(both red) + P(both queens) minus P(both red queens). Substituting the fractions and simplifying leads again to 55 / 221, so the combination approach is confirmed.

Why Other Options Are Wrong:
17/112 and 33/221 do not correspond to the correct count of 330 favourable pairs out of 1326. 5/26 is larger than 55/221 and does not arise from any natural combination count for this situation.

Common Pitfalls:
A frequent error is to add the probabilities of both red and both queens without subtracting the overlap where the pair consists of two red queens. This double counts that case and produces a probability that is too large. Another mistake is to forget that order does not matter and use 52 * 51 as the denominator instead of C(52, 2). Careful use of the union rule and combinations prevents these issues.

Final Answer:
The probability that either both cards are red or both cards are queens is 55/221.