Two cards are drawn at random from a standard deck of 52 playing cards without replacement. What is the probability that either both cards are black or both cards are queens?

Introduction / Context:
This problem tests understanding of compound events in probability and the inclusion exclusion principle. We have two types of favourable outcomes: both cards are black, or both cards are queens. These events overlap when both cards are black queens, so counting must be done carefully to avoid double counting. A standard 52 card deck with equal suits and ranks is assumed.

Given Data / Assumptions:

A standard deck has 52 cards: 26 black (spades and clubs) and 26 red (hearts and diamonds).

There are 4 queens in total, of which 2 are black (queen of spades and queen of clubs).

Two cards are drawn at random without replacement.

Each unordered pair of cards is equally likely.

Concept / Approach:
We consider two events: A = both cards are black, B = both cards are queens. The required probability is P(A or B). For overlapping events, P(A or B) = P(A) + P(B) - P(A and B). We compute each probability using combinations: total possible pairs from 52 cards and favourable pairs for each event. Subtracting the intersection term avoids double counting pairs that satisfy both conditions at once.

Step-by-Step Solution:
Total number of ways to choose 2 cards from 52 = C(52,2).C(52,2) = 52 * 51 / 2 = 1326.Event A: both cards are black.Number of black cards = 26, so ways to choose 2 black cards = C(26,2) = 26 * 25 / 2 = 325.Event B: both cards are queens.Number of queens = 4, so ways to choose 2 queens = C(4,2) = 4 * 3 / 2 = 6.Intersection A and B: both cards are black queens.Number of black queens = 2, so ways to choose both = C(2,2) = 1.Total favourable pairs = 325 + 6 - 1 = 330.Required probability = 330 / 1326.Simplify by dividing numerator and denominator by 6: 330 / 6 = 55 and 1326 / 6 = 221.Thus probability = 55/221.
Verification / Alternative check:
We can cross check by computing P(A) and P(B) separately and subtracting P(A and B).P(A) = 325 / 1326, P(B) = 6 / 1326, P(A and B) = 1 / 1326.Then P(A or B) = P(A) + P(B) - P(A and B) = (325 + 6 - 1) / 1326 = 330 / 1326 = 55/221.This agrees with the direct favourable counting result.
Why Other Options Are Wrong:
52/221 and 55/190 are incorrect simplifications and do not correspond to the correct counting. 19/221 and 5/52 underestimate the probability, likely due to missing some favourable cases or incorrect handling of the overlap. Only 55/221 matches the careful inclusion exclusion calculation.

Common Pitfalls:
A common mistake is to add the counts for both black and both queens without subtracting the intersection, thereby double counting the pair of black queens. Some students also confuse ordered pairs with unordered combinations, leading to incorrect denominators. Remember that in card problems, combinations are usually used because order does not matter when drawing multiple cards at once.

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