20 carefully crafted questions covering the most tested — and most missed — concepts. Pick an answer to get instant feedback & detailed explanations.
Option C is false. {1,2,3,4,5,6} contains the element 6, which is not in A. A subset must have every element inside the original set.
✅ A: 3 ∈ A — 3 is an element of A. TRUE.
✅ B: {2,4} ⊆ A — Both 2 and 4 are in A. TRUE.
❌ C: {1,2,3,4,5,6} ⊆ A — 6 is not in A. FALSE.
✅ D: The empty set ∅ is a subset of every set. TRUE.
A ∩ B = elements that appear in both A and B.
Check each element of A: 2 (not in B), 4 ✓, 6 (not in B), 8 ✓.
A ∩ B = {4, 8}
Option A is A ∪ B (union). Option C is neither union nor intersection. Option D has duplicates — sets never repeat elements.
The complement A' = U minus A = all elements in U that are NOT in A.
U = {1,2,3,4,5,6,7,8,9,10}, A = even numbers = {2,4,6,8,10}.
A' = {1,3,5,7,9} (the odd numbers)
❌ B = A itself. ❌ C = ∅ would only be true if A = U. ❌ D includes 0, which is not in U.
Apply the Inclusion-Exclusion Principle:
n(F ∪ B) = n(F) + n(B) − n(F ∩ B)
= 18 + 15 − 7 = 26
❌ A (33): forgot to subtract the overlap — you counted "both" people twice.
❌ B (25): common arithmetic slip.
❌ D (30): that's the total class size, not the answer.
For a set with n = 3 elements, the number of subsets = 2³ = 8.
∅, {a}, {b}, {c}, {a,b}, {a,c}, {b,c}, {a,b,c}
❌ B (7): forgot to count the empty set ∅ — it's always a subset!
❌ A (6): forgot both ∅ and the full set.
Rule: each element has 2 choices (in or out) → 2 × 2 × 2 = 8.
Total marbles = 3 + 4 + 5 = 12. Blue marbles = 4.
P(blue) = 4/12 = 1/3
❌ A (1/4): incorrectly used 3 in the denominator instead of 12.
❌ C (4/8): wrong total.
❌ D (5/12): that's the probability of green.
Method: complement.
P(no 6 on first roll) = 5/6
P(no 6 on second roll) = 5/6
P(no 6 on either) = 5/6 × 5/6 = 25/36
P(at least one 6) = 1 − 25/36 = 11/36
❌ A: P(6 on both rolls) = 1/36 — that's "exactly two 6s".
❌ B: 2/6 = 12/36 — double-counted the case (6,6).
A card cannot be both a King and a Queen simultaneously → A ∩ B = ∅ → Mutually exclusive.
P(King) = 4/52 P(Queen) = 4/52
P(A ∪ B) = 4/52 + 4/52 = 8/52 = 2/13
❌ B: wrong — they ARE mutually exclusive.
❌ C: 4/52 only counts Kings or only Queens, not both.
Coin flip and die roll are independent events.
P(Heads) = 1/2
P(rolling a 4) = 1/6
P(Heads AND 4) = 1/2 × 1/6 = 1/12
❌ A (1/4): used wrong denominator for the die.
❌ C (1/6): forgot to multiply by P(Heads).
❌ D: this would be an addition/union calculation, not "and".
Conditional probability formula:
P(A | B) = P(A ∩ B) ÷ P(B) = 0.2 ÷ 0.4 = 0.5
Notice: P(A|B) = P(A) = 0.5 → This means A and B are actually independent! (Knowing B happened doesn't change the probability of A.)
❌ A (0.2): that's P(A ∩ B), not the conditional probability.
❌ D (0.8): that would be P(B|A) formula applied incorrectly.
5! = 5 × 4 × 3 × 2 × 1
= 20 × 6 = 120
❌ A (60): stopped at 5 × 4 × 3 = 60 without multiplying by 2 × 1.
❌ C (25): just 5² — not factorial.
❌ D (720): that's 6! = 6 × 5 × 4 × 3 × 2 × 1.
C(7,3) = 7! / (3! × 4!)
= (7 × 6 × 5) / (3 × 2 × 1)
= 210 / 6 = 35
❌ A (21): that's C(7,2).
❌ B (42): arithmetic error (forgot to divide by 3!).
❌ D (210): that's the numerator only — 7 × 6 × 5 — didn't divide by 3!.
This is a permutation (order matters = arrangement). P(5,3) = 5! / (5−3)! = 5! / 2! = 120 / 2 = 60 Or think step by step: 1st position → 5 choices, 2nd → 4, 3rd → 3. 5 × 4 × 3 = 60 ❌ B (10): that's C(5,2). ❌ C (120) = 5! — that's arranging all 5. ❌ D (30): arithmetic error.
By the Fundamental Counting Principle, multiply the number of choices at each stage:
4 × 6 × 3 = 72
❌ A (13): added instead of multiplied (4 + 6 + 3).
❌ B (36): forgot starters (6 × 3 × 2).
❌ C (48): 4 × 6 × 2 — wrong dessert count.
Symmetry: C(8,5) = C(8, 8−5) = C(8,3) — much easier to calculate! C(8,3) = (8 × 7 × 6) / (3 × 2 × 1) = 336 / 6 = 56 ❌ A (40): arithmetic error. ❌ C (28): that's C(8,2). ❌ D (336): numerator only, forgot to divide by 3! = 6.
C(4,2) = 6 (ways to choose 2 red from 4)
C(10,2) = 45 (ways to choose any 2 from 10)
P = 6/45 = 2/15
❌ B (4/25): used (4/10)² — that's WITH replacement, not without!
❌ C (1/5): 9/45 = 1/5 → arithmetic error in C(4,2).
❌ D (2/5): miscounted the combinations.
n(T ∪ C) = n(T) + n(C) − n(T ∩ C) = 60 + 50 − 30 = 80
n(neither) = 100 − 80 = 20
❌ A (10): forgot to subtract intersection when computing union.
❌ C (30): that's n(T ∩ C), the overlap.
❌ D (40): used 100 − 60 (only subtracted tea drinkers).
The term with x²y³ has k = 3 (the power of y). By the Binomial Theorem:
coefficient = C(5,3) = C(5,2) = 10
Row 5 of Pascal's Triangle: 1, 5, 10, 10, 5, 1.
❌ A (5): that's C(5,1) — power of y would be 1. ❌ B (20): no such entry in row 5. ❌ D (15): that's C(6,2).
Step 1: Treat AB as one unit → 4 units total: [AB], C, D, E.
4! = 24 arrangements of 4 units
Step 2: Inside the unit, A and B can swap: AB or BA.
2! = 2 internal arrangements
Total = 4! × 2! = 24 × 2 = 48
❌ A (24): forgot that AB can also be arranged as BA.
❌ C (120): 5! — no restriction applied.
❌ D (72): over-counted.
E(winnings) = (1/6) × $10 + (5/6) × $0 = $10/6 ≈ $1.667
E(profit) = $1.667 − $2.00 = −$0.333 ≈ −$0.33
The game has a negative expected profit — you lose about $0.33 per play on average. This is why casinos make money!
❌ A ($1.67): forgot to subtract the $2 ticket cost.
❌ D ($8): $10 − $2 = $8 but forgot to multiply by probability 1/6.
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