Inclusion–Exclusion for Two Sets (Applied & Algebraic Focus)

P(A∪B) = 0.4 + 0.5 - 0.1 = 0.8, confirming A. Exactly-one = 0.4 + 0.5 - 0.2 = 0.7, confirming B. The formula for exactly one is P(A) + P(B) - 2P(A∩B), confirming C. Option D is false because P(A∩B) = 0.1, not 0.

The intersection contains exactly the elements belonging to both sets simultaneously. Option A describes exclusive elements. Option B describes elements in neither set. Option D describes elements outside the union entirely.

The answer is True. A∩B = 10 + 8 - 15 = 3. Three elements belong to both sets, which is why the union falls below the sum of 18.

P(A∪B) = 0.3 + 0.6 - 0.2 = 0.7. The 0.2 probability of both events is subtracted once to avoid double-counting. Option A gives 0.5, option B gives 0.6, option D gives 0.9.

The answer is True. All three structures obey the same union-intersection relationship. For sets it governs element counts, for probabilities it governs likelihoods, and for logic events it governs compound outcome probabilities.

The intersection is subtracted because adding A and B counts shared elements twice. Subtracting the intersection once removes the double count, giving the true number of distinct elements in the union.

The answer is True. The formula uses A + B - A∩B. Swapping A and B gives B + A - B∩A, which equals the same value since addition and intersection are both commutative.

Disjoint sets share no members so their intersection is the empty set, which has size 0. Option A implies one shared element. Option B implies two. Option D applies when one set is fully contained in the other.

Union = 28 + 25 - 9 = 44, confirming A. Exactly-one = (28-9) + (25-9) = 19 + 16 = 35, confirming B. Option C states intersection = 0 but it is given as 9. Option D is false because the intersection equals 9.

The answer is False. Two sets of equal size can share any number of elements from 0 up to the full size. For example A = {1,2,3} and B = {4,5,6} have equal sizes but share nothing. Equal sizes place no constraint on overlap.

Union = 60 + 45 - 25 = 80, confirming B. Exactly-one = (60-25) + (45-25) = 35 + 20 = 55, confirming C. Option A adds without subtracting the overlap. Option D is false because the intersection equals 25, so the sets share members and are not disjoint.

The answer is True. With intersection = 0 the formula becomes union = A + B - 0 = A + B. There are no shared elements to subtract so every element is counted exactly once.

Union = (x+3) + 2x - x = 2x + 3, confirming A. Exactly-one = (x+3-x) + (2x-x) = 3 + x = x+3, confirming B. The intersection equals x as given, confirming C. Option D is false because the intersection equals x which can be greater than 0.

A∩B = 70 + 50 - 100 = 20. Since the total of 120 exceeds the class size of 100, at least 20 students must be in both groups. Option A gives 15, option B gives 18, option D gives 25.

Union = 70 + 50 - 20 = 100, confirming A. Exactly-one = (70-20) + (50-20) = 50 + 30 = 80, confirming B. The intersection is given as 20, confirming C. Option D is false because 20 students take both subjects.

The answer is True. The formula gives union = A + B - intersection, and since the intersection is always non-negative, subtracting it cannot increase the result above A + B. If the union exceeds A + B, an arithmetic error occurred.

A∩B = 30 + 35 - 50 = 15, confirming A. Exactly-one = (30-15) + (35-15) = 15 + 20 = 35, confirming B. Option C ignores the overlap entirely. Option D is false because the intersection equals 15.

Union = 25 + 20 - 5 = 40. Without subtracting the overlap the sum would be 45, overcounting the 5 shared elements. Option A gives 35, option B gives 38, option D gives 42.

Union = 40 + 35 - 15 = 60, confirming A. Exactly-one = (40-15) + (35-15) = 25 + 20 = 45, confirming B. Rearranging the union formula gives A∩B = A + B - union, confirming C. Option D is false because the intersection equals 15.

The answer is False. P(A∩B) = 0.5 + 0.4 - 0.7 = 0.2. Since the intersection probability is 0.2 which is greater than 0, the events share outcomes and are not disjoint. Disjoint events require P(A∩B) = 0.