Sets, Probability
& Combinations

💡 Solution \( A \cup B \) means everything in A or B (or both). Combine all elements without repeating: \(\{1,2,3,4,6,8\}\). Common mistake: choosing only the shared elements — that's \(A \cap B = \{2,4\}\), not the union!

💡 Solution \( A' = U \setminus A \) — remove every element of \(A\) from \(U\). \(U = \{1..8\}\), \(A = \{2,4,6,8\}\), so what's left is \(\{1,3,5,7\}\). Trick: students sometimes mix up \(A'\) with \(B'\). Always refer back to \(U\).

💡 Solution Step 1 — find \(|A \cup B| = 18 + 12 - 5 = 25\). Step 2 — neither = total \(-\) union \(= 30 - 25 = 5\). The classic trap: forgetting to subtract the overlap, giving 30 wrong!

💡 Solution Every element of \(A = \{1,3,5\}\) appears in \(B\), so \(A \subseteq B\). But \(B\) has 2 and 4 which aren't in \(A\), so \(B \not\subseteq A\). Students often reverse the direction of the arrow!

💡 Solution \(A \cup B = \{1,2,3,4,5,6,7\}\). Its complement in \(U\) is what's not in the union: \(\{8,9,10\}\). Alternatively by De Morgan's Law: \(A' = \{6..10\}\), \(B' = \{1,2,3,8,9,10\}\), and \(A' \cap B' = \{8,9,10\}\). ✓

💡 Solution Total = 12. Not red = \(3+5 = 8\). So \(P(\text{not red}) = \frac{8}{12} = \frac{2}{3}\). Or use complement: \(1 - P(\text{red}) = 1 - \frac{4}{12} = \frac{2}{3}\). Shortcut: P(A') = 1 − P(A).

💡 Solution Mutually exclusive means they can't happen together, so \(P(A \cap B) = 0\). Therefore \(P(A \cup B) = P(A) + P(B) = 0.3 + 0.45 = 0.75\). Trap: 0.135 = \(P(A) \times P(B)\) — that's for independent events, not this!

💡 Solution We want \(P(\text{Spanish} \mid \text{French}) = \frac{P(\text{both})}{P(\text{French})} = \frac{8/50}{30/50} = \frac{8}{30} = \frac{4}{15}\). Key: the denominator shrinks to the condition group (French = 30), not the whole 50.

💡 Solution \(P(\text{heads}) = \frac{1}{2}\). Numbers \(> 4\) on a die: \{5, 6\}, so \(P = \frac{2}{6} = \frac{1}{3}\). Since independent: \(P = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}\). Trap: answer D (\(\frac{1}{12}\)) misreads "greater than 4" as only \{5\}.

💡 Solution \(P(A \cup B) = 0.5 + 0.4 - 0.2 = 0.7\). The overlap is subtracted once because it's counted in both \(P(A)\) and \(P(B)\). Forgetting to subtract gives 0.9 — the most common error on this type.

💡 Solution \(\binom{7}{3} = \frac{7!}{3! \cdot 4!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = \frac{210}{6} = 35\). Trap: 210 = \(7 \times 6 \times 5\) — that's the permutation without dividing. Always divide by \(r!\) when order doesn't matter.

💡 Solution \(_{6}P_{4} = \frac{6!}{(6-4)!} = \frac{6!}{2!} = 6 \times 5 \times 4 \times 3 = 360\). Order matters here (a line). Trap: 15 = \(\binom{6}{4}\) — that's combinations (no order). 720 = \(6!\) — that's arranging all 6.

💡 Solution Multiply independent choices: \(4 \times 5 \times 3 = 60\). Each course is an independent step. Trap: students add (getting 12) instead of multiply. The multiplication principle applies when choosing one from each group.

💡 Solution Choose 2 girls from 5: \(\binom{5}{2} = 10\). Choose 2 boys from 4: \(\binom{4}{2} = 6\). Multiply: \(10 \times 6 = 60\). Since girl-choices and boy-choices are independent groups, we multiply — not add. This is the key insight for split-group problems.

💡 Solution With repeated elements, divide total arrangements by the factorial of each repeat count. S appears 3 times, T appears 3 times, I appears 2 times: \(\frac{10!}{3! \cdot 3! \cdot 2!} = 50400\). Students often forget one repeated group — always count every repeated letter.

💡 Solution \(P(\text{1st red}) = \frac{3}{5}\). After removing one red, bag has 2 red, 2 blue (4 total). \(P(\text{2nd red}) = \frac{2}{4} = \frac{1}{2}\). Multiply: \(\frac{3}{5} \times \frac{1}{2} = \frac{3}{10}\). Trap: answer B uses \(\frac{3}{5} \times \frac{3}{5}\) — that's WITH replacement!

💡 Solution In A or B: \(12+8+10 = 30\). Neither: \(40 - 30 = 10\). So \(P(A' \cap B') = \frac{10}{40} = \frac{1}{4}\). \(A' \cap B'\) is the "outside both circles" region — a very common exam wording. Don't confuse it with \(A \cap B = \{8\}\).

💡 Solution Choose exactly 2 aces from 4: \(\binom{4}{2}\). The remaining 3 cards must be non-aces from 48: \(\binom{48}{3}\). Total hands: \(\binom{52}{5}\). Answer: \(\frac{\binom{4}{2}\binom{48}{3}}{\binom{52}{5}}\). Trap C picks 2 non-aces instead of 3 — always check: 2 aces + 3 others = 5 cards total. ✓

💡 Solution Use complement: \(P(\text{at least one 6}) = 1 - P(\text{no 6 at all})\). \(P(\text{no 6 on one roll}) = \frac{5}{6}\). Three independent rolls: \(P(\text{no 6}) = \left(\frac{5}{6}\right)^3 = \frac{125}{216}\). Therefore: \(1 - \frac{125}{216} = \frac{91}{216}\). Direct counting is much harder — always use complement for "at least one"!

💡 Solution Find \(P(A \cap B)\): use addition rule → \(0.8 = 0.6 + 0.5 - P(A \cap B)\) → \(P(A \cap B) = 0.3\). Check independence: \(P(A) \cdot P(B) = 0.6 \times 0.5 = 0.30\). Since \(0.3 = 0.30\), they ARE independent! This is a beautifully tricky question — work through the arithmetic carefully, and the answer surprises you.