Trick: "Not happening" = subtract from 1.
Total marbles = 3 + 5 + 2 = 10. Blue marbles = 5, so \(P(\text{blue}) = \frac{5}{10} = \frac{1}{2}\).
Complement rule: \(P(\text{not blue}) = 1 - \frac{1}{2} = \frac{1}{2}\). The red and green together = 5 marbles, so \(\frac{5}{10} = \frac{1}{2}\). ✓
Sample space shrinks! Even outcomes = {2, 4, 6} — only 3 possibilities.
Among these, greater than 3: {4, 6} — that's 2 outcomes.
\(P(\text{> 3} \mid \text{even}) = \dfrac{2}{3}\) ✓
Common mistake: Using \(\frac{2}{6}\) — this ignores the condition "given even".
Addition Rule: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
\(= 0.4 + 0.5 - 0.2 = \mathbf{0.7}\) ✓
Trap: Option A (0.9) forgets to subtract the intersection. Option D (1.1) adds everything — probability can never exceed 1!
Formula: \(\binom{n}{r} = \dfrac{n!}{r!(n-r)!}\) · Keyword: "choose", "select", "group" → combination.
Keyword: "arrange", "order", "rank" → permutation.
"Choose" = Combination, order doesn't matter.
\(\binom{8}{3} = \dfrac{8!}{3! \cdot 5!} = \dfrac{8 \times 7 \times 6}{3 \times 2 \times 1} = \dfrac{336}{6} = \mathbf{56}\) ✓
Option B (336) = \(_8P_3\) = permutation — this would be the answer if order mattered.
Glue trick: Treat Ali+Ben as one unit → 4 units total.
Arrange 4 units: \(4! = 24\) ways.
Ali and Ben can swap inside their unit: \(2! = 2\) ways.
Total: \(24 \times 2 = \mathbf{48}\) ✓
Option C (120 = 5!) ignores the constraint entirely.
Choose 2 women from 5: \(\binom{5}{2} = 10\)
Choose 2 men from 6: \(\binom{6}{2} = 15\)
Multiply (independent choices): \(10 \times 15 = \mathbf{150}\) ✓
Option B (330) = \(\binom{11}{4}\) = total unrestricted committees — ignores the "exactly 2 women" condition.
Symmetry: \(\binom{n}{r} = \binom{n}{n-r}\), so \(\binom{10}{7} = \binom{10}{3}\).
\(\binom{10}{3} = \dfrac{10 \times 9 \times 8}{6} = 120\)
Therefore: \(120 + 120 = \mathbf{240}\) ✓
Key insight: \(\binom{10}{3} = \binom{10}{7}\) — choosing 3 to include is the same as choosing 7 to exclude.
Conditional: P(A|B) = P(A∩B) / P(B)
Test for independence: Check if \(P(A|B) = P(A)\). If yes → independent. If no → dependent.
Independent events: \(P(A \cap B) = P(A) \times P(B)\)
\(= 0.3 \times 0.4 = \mathbf{0.12}\) ✓
You can ONLY multiply like this when events are independent. This is the definition of independence.
Independence test: \(P(A) \times P(B) = 0.6 \times 0.5 = 0.30\).
Since \(P(A \cap B) = 0.3 = P(A) \times P(B)\), the events ARE independent. ✓
Option A is a very common mistake: \(P(A \cap B) = 0\) means mutually exclusive, NOT dependent.
Conditional probability formula: \(P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}\)
\(= \dfrac{0.12}{0.4} = \mathbf{0.30}\) ✓
Trick: "Given B happened" → B becomes your new sample space. Divide the intersection by P(B).
\(P(\text{1st red}) = \dfrac{4}{10}\). After removing one red ball: 3 red left, 9 total.
\(P(\text{2nd red} \mid \text{1st red}) = \dfrac{3}{9}\)
\(P(\text{both}) = \dfrac{4}{10} \times \dfrac{3}{9} = \dfrac{12}{90} = \dfrac{2}{15}\) ✓
Option A uses \(\frac{4}{10} \times \frac{4}{10}\) — that's WITH replacement (wrong here).
Mutually exclusive: \(P(A \cap B) = 0\) → cannot happen together.
Exhaustive: \(P(A \cup B) = 1\) → together they cover everything.
\(|F \cup B| = |F| + |B| - |F \cap B| = 18 + 15 - 7 = 26\)
Neither = Total \(-\) Union \(= 30 - 26 = \mathbf{4}\) ✓
Common mistake: Adding 18 + 15 = 33 (> 30!), forgetting to subtract the overlap.
"C only" = coffee but NOT tea = \(30 - 10 = 20\) people.
\(P(\text{C only}) = \dfrac{20}{50} = 0.4\) ✓
Option B gives P(C) which includes those who like both — that's not "only".
De Morgan's Law: \(A' \cap B' = (A \cup B)'\)
Since mutually exclusive: \(P(A \cup B) = 0.35 + 0.45 = 0.80\)
\(P((A \cup B)') = 1 - 0.80 = \mathbf{0.20}\) ✓
Option C multiplies \(P(A') \times P(B')\) — only valid if A and B are independent, which they are NOT (mutually exclusive ≠ independent).
With replacement → multiply same fractions
Without replacement → denominator decreases
"At least one" → use complement: \(1 - P(\text{none})\)
Complement trick: \(P(\text{at least 1}) = 1 - P(\text{none})\)
\(P(\text{no rain day 1}) = 0.7\), \(P(\text{no rain day 2}) = 0.7\)
\(P(\text{no rain either day}) = 0.7 \times 0.7 = 0.49\)
\(P(\text{at least 1 rainy}) = 1 - 0.49 = \mathbf{0.51}\) ✓
Option B: \(0.3 + 0.3 = 0.6\) — double-counting the day it rains on both!
Permutation: \(_5P_4 = \dfrac{5!}{(5-4)!} = \dfrac{5!}{1!} = 5 \times 4 \times 3 \times 2 = \mathbf{120}\) ✓
Or: 5 choices × 4 × 3 × 2 = 120.
Option C (625 = 5⁴) allows repetition — wrong here.
Option D (24 = 4!) treats it as arranging 4 fixed items.
Kings = 4, Hearts = 13. But King of Hearts is counted in BOTH!
Addition rule: \(P(K \cup H) = \dfrac{4}{52} + \dfrac{13}{52} - \dfrac{1}{52} = \dfrac{16}{52} = \dfrac{4}{13}\) ✓
Option A gives \(\frac{17}{52}\) — forgetting to subtract the overlap (King of Hearts).
Circular permutations: fix one person, arrange the rest.
\((n-1)! = (8-1)! = 7! = \mathbf{5040}\) ✓
Option A \((8!)\) treats it as a line — but rotations of the same arrangement look identical at a round table, so we divide by \(n = 8\), giving \(\frac{8!}{8} = 7!\).
Rearrange conditional formula: \(P(A \cap B) = P(A \mid B) \times P(B)\)
\(= 0.6 \times 0.5 = \mathbf{0.30}\) ✓
This is the multiplication rule for dependent events. Remember: the conditional formula can be solved for the intersection.
Repetition allowed: each letter position has 26 choices, each digit has 10.
Total \(= 26 \times 26 \times 26 \times 10 \times 10 = 26^3 \times 10^2 = 17{,}576 \times 100 = \mathbf{1{,}757{,}600}\) ✓
Option B: No repetition formula (wrong — question says repeats allowed).
Option C: Combination ignores order — but "AAB" ≠ "ABA" for passwords!
Option D: Adding instead of multiplying — never add independent sequential choices.