{
  "deckName": "Exam P — Set Theory & Combinatorics",
  "examCode": "Exam P",
  "cards": [
    {
      "front": "What is a **sample space** $S$ and an **event** in a probability model?",
      "back": "The sample space $S$ is the set of all possible outcomes of a random experiment. An event is any subset $A\\subseteq S$. A single outcome $\\omega\\in S$ is an elementary (or simple) event, and $S$ itself is the certain event while $\\emptyset$ is the impossible event.",
      "tag": "Sample spaces & events"
    },
    {
      "front": "Define the three set operations **union**, **intersection**, and **complement** for events $A,B\\subseteq S$.",
      "back": "Union: $A\\cup B=\\{\\omega:\\omega\\in A \\text{ or } \\omega\\in B\\}$ (at least one occurs).\nIntersection: $A\\cap B=\\{\\omega:\\omega\\in A \\text{ and } \\omega\\in B\\}$ (both occur).\nComplement: $A^{c}=\\{\\omega\\in S:\\omega\\notin A\\}$ (A does not occur).",
      "tag": "Set operations"
    },
    {
      "front": "State **De Morgan's laws** for two events.",
      "back": "$(A\\cup B)^{c}=A^{c}\\cap B^{c}$ and $(A\\cap B)^{c}=A^{c}\\cup B^{c}$. In words: the complement of a union is the intersection of the complements, and vice versa. They generalize to $\\left(\\bigcup_{i}A_{i}\\right)^{c}=\\bigcap_{i}A_{i}^{c}$.",
      "tag": "Set operations"
    },
    {
      "front": "What does it mean for events $A$ and $B$ to be **mutually exclusive** (disjoint)?",
      "back": "They cannot both occur, i.e. $A\\cap B=\\emptyset$, so $\\Pr(A\\cap B)=0$. For disjoint events the addition rule simplifies to $\\Pr(A\\cup B)=\\Pr(A)+\\Pr(B)$. Disjoint is not the same as independent.",
      "tag": "Set operations"
    },
    {
      "front": "State the three **Kolmogorov axioms** of probability.",
      "back": "1) Non-negativity: $\\Pr(A)\\geq 0$ for every event $A$.\n2) Normalization: $\\Pr(S)=1$.\n3) Countable additivity: for pairwise disjoint $A_{1},A_{2},\\ldots$, $\\Pr\\!\\left(\\bigcup_{i=1}^{\\infty}A_{i}\\right)=\\sum_{i=1}^{\\infty}\\Pr(A_{i})$.",
      "tag": "Axioms"
    },
    {
      "front": "Using the axioms, derive the **complement rule** and the value of $\\Pr(\\emptyset)$.",
      "back": "Since $A$ and $A^{c}$ are disjoint with $A\\cup A^{c}=S$, additivity gives $\\Pr(A)+\\Pr(A^{c})=\\Pr(S)=1$, so $\\Pr(A^{c})=1-\\Pr(A)$. Taking $A=S$ gives $\\Pr(\\emptyset)=1-\\Pr(S)=0$.",
      "tag": "Axioms"
    },
    {
      "front": "If $A\\subseteq B$, what can you say about $\\Pr(A)$ versus $\\Pr(B)$, and why?",
      "back": "Monotonicity: $\\Pr(A)\\leq\\Pr(B)$. Write $B=A\\cup(B\\cap A^{c})$ as a disjoint union, so $\\Pr(B)=\\Pr(A)+\\Pr(B\\cap A^{c})\\geq\\Pr(A)$ because the second term is $\\geq 0$. A corollary is that every probability satisfies $0\\leq\\Pr(A)\\leq 1$.",
      "tag": "Axioms"
    },
    {
      "front": "State the **addition rule** (inclusion–exclusion for two events) $\\Pr(A\\cup B)$.",
      "back": "$\\Pr(A\\cup B)=\\Pr(A)+\\Pr(B)-\\Pr(A\\cap B)$. The intersection is subtracted once because it is counted in both $\\Pr(A)$ and $\\Pr(B)$. If $A\\cap B=\\emptyset$ the last term is $0$.",
      "tag": "Inclusion-exclusion"
    },
    {
      "front": "State **inclusion–exclusion** for three events $\\Pr(A\\cup B\\cup C)$.",
      "back": "$\\Pr(A\\cup B\\cup C)=\\Pr(A)+\\Pr(B)+\\Pr(C)-\\Pr(A\\cap B)-\\Pr(A\\cap C)-\\Pr(B\\cap C)+\\Pr(A\\cap B\\cap C)$. Pattern: add singles, subtract pairwise intersections, add back the triple intersection.",
      "tag": "Inclusion-exclusion"
    },
    {
      "front": "In a group, $\\Pr(A)=0.5$, $\\Pr(B)=0.4$, and $\\Pr(A\\cap B)=0.2$. Find $\\Pr(A\\cup B)$ and $\\Pr(A^{c}\\cap B^{c})$.",
      "back": "Addition rule: $\\Pr(A\\cup B)=0.5+0.4-0.2=0.7$.\nBy De Morgan, $\\Pr(A^{c}\\cap B^{c})=\\Pr\\!\\big((A\\cup B)^{c}\\big)=1-\\Pr(A\\cup B)=1-0.7=0.30$.",
      "tag": "Inclusion-exclusion"
    },
    {
      "front": "An insurer's policyholders carry coverages $A$, $B$, $C$ with $\\Pr(A)=\\Pr(B)=\\Pr(C)=0.30$, each pairwise intersection $=0.10$, and $\\Pr(A\\cap B\\cap C)=0.05$. What fraction carry **at least one** coverage?",
      "back": "By inclusion–exclusion, $\\Pr(A\\cup B\\cup C)=3(0.30)-3(0.10)+0.05=0.90-0.30+0.05=0.65$. So $65\\%$ carry at least one coverage (and $35\\%$ carry none).",
      "tag": "Inclusion-exclusion"
    },
    {
      "front": "With the coverages from the previous card ($\\Pr$ each $=0.30$, pairwise $=0.10$, triple $=0.05$), what fraction carry **exactly one** coverage?",
      "back": "The 'exactly one' count for three events is $\\sum\\Pr(\\text{single})-2\\sum\\Pr(\\text{pairwise})+3\\Pr(\\text{triple})$.\n$=3(0.30)-2(3)(0.10)+3(0.05)=0.90-0.60+0.15=0.45$. So $45\\%$ carry exactly one coverage.",
      "tag": "Inclusion-exclusion"
    },
    {
      "front": "When are outcomes **equally likely**, and how is $\\Pr(A)$ computed in that case?",
      "back": "When $S$ is finite and every outcome has the same probability $1/|S|$. Then for any event $A$, $\\Pr(A)=\\frac{|A|}{|S|}=\\frac{\\text{number of favorable outcomes}}{\\text{total number of outcomes}}$. This reduces probability to a counting problem.",
      "tag": "Equally likely & counting"
    },
    {
      "front": "How many **arrangements** (orderings) are there of $n$ distinct objects, and why?",
      "back": "$n!=n\\cdot(n-1)\\cdots 2\\cdot 1$. The first position has $n$ choices, the next $n-1$, and so on by the multiplication principle, giving the product $n!$. By convention $0!=1$.",
      "tag": "Permutations"
    },
    {
      "front": "Define the number of **permutations** of $r$ objects chosen from $n$ distinct objects, ${}_{n}P_{r}$.",
      "back": "${}_{n}P_{r}=\\frac{n!}{(n-r)!}=n(n-1)\\cdots(n-r+1)$. Order matters: it counts ordered selections (arrangements) of size $r$ from $n$ distinct items without replacement.",
      "tag": "Permutations"
    },
    {
      "front": "Define the **combination** (binomial coefficient) $\\binom{n}{r}$ and state its symmetry property.",
      "back": "$\\binom{n}{r}=\\frac{n!}{r!\\,(n-r)!}=\\frac{{}_{n}P_{r}}{r!}$, the number of unordered subsets of size $r$ from $n$ distinct items. Symmetry: $\\binom{n}{r}=\\binom{n}{n-r}$. Also $\\binom{n}{0}=\\binom{n}{n}=1$.",
      "tag": "Combinations"
    },
    {
      "front": "How many ways can a committee of $4$ be chosen from $10$ people? Compute the number.",
      "back": "Order does not matter, so use a combination: $\\binom{10}{4}=\\frac{10!}{4!\\,6!}=\\frac{10\\cdot 9\\cdot 8\\cdot 7}{4\\cdot 3\\cdot 2\\cdot 1}=\\frac{5040}{24}=210$.",
      "tag": "Combinations"
    },
    {
      "front": "A bag has $6$ red and $4$ blue chips. Draw $3$ without replacement. Find $\\Pr(\\text{exactly }2\\text{ red})$.",
      "back": "Equally likely subsets of size $3$: total $\\binom{10}{3}=120$. Favorable: choose $2$ reds and $1$ blue, $\\binom{6}{2}\\binom{4}{1}=15\\cdot 4=60$.\n$\\Pr=\\frac{60}{120}=0.50$.",
      "tag": "Equally likely & counting"
    },
    {
      "front": "How many distinct arrangements of the letters in **MISSISSIPPI** are there?",
      "back": "There are $11$ letters with repeats $\\text{M}{:}1,\\ \\text{I}{:}4,\\ \\text{S}{:}4,\\ \\text{P}{:}2$. Use the multinomial (permutations of a multiset):\n$\\frac{11!}{1!\\,4!\\,4!\\,2!}=\\frac{39{,}916{,}800}{1\\cdot 24\\cdot 24\\cdot 2}=\\frac{39{,}916{,}800}{1152}=34{,}650$.",
      "tag": "Multinomial & partitions"
    },
    {
      "front": "State the **multinomial coefficient** that counts arrangements of $n$ items in $k$ groups of sizes $n_{1},\\ldots,n_{k}$ (with $\\sum n_{i}=n$).",
      "back": "$\\binom{n}{n_{1},n_{2},\\ldots,n_{k}}=\\frac{n!}{n_{1}!\\,n_{2}!\\cdots n_{k}!}$. It counts the ways to partition $n$ distinct objects into labeled groups of the given sizes (equivalently, distinct orderings of a multiset). When $k=2$ this reduces to $\\binom{n}{n_{1}}$.",
      "tag": "Multinomial & partitions"
    },
    {
      "front": "In how many ways can $12$ distinct claims be split into three **labeled** teams of sizes $5$, $4$, and $3$? Compute it.",
      "back": "Use the multinomial coefficient: $\\frac{12!}{5!\\,4!\\,3!}=\\frac{479{,}001{,}600}{120\\cdot 24\\cdot 6}=\\frac{479{,}001{,}600}{17{,}280}=27{,}720$.",
      "tag": "Multinomial & partitions"
    },
    {
      "front": "How many ways are there to partition $9$ distinct objects into three **unlabeled** groups of $3$ each? Compute it.",
      "back": "First count labeled groups: $\\frac{9!}{3!\\,3!\\,3!}=\\frac{362{,}880}{216}=1680$. Because the three groups are the same size and **unlabeled**, divide by $3!=6$ for the interchangeable group labels:\n$\\frac{1680}{6}=280$.",
      "tag": "Multinomial & partitions"
    },
    {
      "front": "State the **binomial theorem** for $(x+y)^{n}$.",
      "back": "$(x+y)^{n}=\\displaystyle\\sum_{k=0}^{n}\\binom{n}{k}x^{k}y^{n-k}$. The coefficient of the $x^{k}y^{n-k}$ term is the binomial coefficient $\\binom{n}{k}$, the number of ways to choose which $k$ of the $n$ factors contribute an $x$.",
      "tag": "Binomial theorem"
    },
    {
      "front": "Use the binomial theorem to evaluate $\\displaystyle\\sum_{k=0}^{n}\\binom{n}{k}$ and $\\displaystyle\\sum_{k=0}^{n}(-1)^{k}\\binom{n}{k}$.",
      "back": "Set $x=y=1$ in $(x+y)^{n}$: $\\sum_{k=0}^{n}\\binom{n}{k}=2^{n}$ (total number of subsets of an $n$-set).\nSet $x=-1,\\ y=1$: $\\sum_{k=0}^{n}(-1)^{k}\\binom{n}{k}=(-1+1)^{n}=0$ for $n\\geq 1$.",
      "tag": "Binomial theorem"
    },
    {
      "front": "Find the coefficient of $x^{3}$ in the expansion of $(2x+3)^{5}$.",
      "back": "General term: $\\binom{5}{k}(2x)^{k}(3)^{5-k}$. For $x^{3}$ take $k=3$:\n$\\binom{5}{3}(2)^{3}(3)^{2}=10\\cdot 8\\cdot 9=720$. So the coefficient of $x^{3}$ is $720$.",
      "tag": "Binomial theorem"
    },
    {
      "front": "**Pascal's rule:** express $\\binom{n}{k}$ in terms of $\\binom{n-1}{\\cdot}$ coefficients.",
      "back": "$\\binom{n}{k}=\\binom{n-1}{k-1}+\\binom{n-1}{k}$. Interpretation: of the $n$ items, fix one special item; subsets of size $k$ either contain it ($\\binom{n-1}{k-1}$ ways) or do not ($\\binom{n-1}{k}$ ways). This is the rule that builds Pascal's triangle.",
      "tag": "Binomial theorem"
    },
    {
      "front": "Five people are seated in a row at random. Find $\\Pr(\\text{two specific people sit together})$.",
      "back": "Total arrangements: $5!=120$. Glue the two specified people into one block (treat as a single unit): $4!$ orders of the $4$ units, times $2!$ internal orders $=24\\cdot 2=48$ favorable.\n$\\Pr=\\frac{48}{120}=0.40$.",
      "tag": "Permutations"
    },
    {
      "front": "A standard $52$-card deck deals a $5$-card hand. Find $\\Pr(\\text{four of a kind})$.",
      "back": "Total hands: $\\binom{52}{5}=2{,}598{,}960$. Count four-of-a-kind hands: choose the rank that appears four times ($13$ ways), then the $5$th card from the remaining $48$ cards.\nFavorable $=13\\cdot 48=624$, so $\\Pr=\\frac{624}{2{,}598{,}960}\\approx 0.000240$.",
      "tag": "Equally likely & counting"
    },
    {
      "front": "From a lot of $20$ items, $5$ are defective. A sample of $4$ is drawn without replacement. Find $\\Pr(\\text{exactly one defective})$.",
      "back": "Total samples: $\\binom{20}{4}=4845$. Favorable: $1$ defective and $3$ good, $\\binom{5}{1}\\binom{15}{3}=5\\cdot 455=2275$.\n$\\Pr=\\frac{2275}{4845}\\approx 0.4696$.",
      "tag": "Equally likely & counting"
    },
    {
      "front": "In a room of $5$ people, find the probability that **at least two share a birthday** (ignore leap years; assume $365$ equally likely days).",
      "back": "Use the complement (all distinct):\n$\\Pr(\\text{all distinct})=\\frac{365\\cdot 364\\cdot 363\\cdot 362\\cdot 361}{365^{5}}\\approx 0.97286$.\n$\\Pr(\\text{at least one match})=1-0.97286\\approx 0.0271$.",
      "tag": "Equally likely & counting"
    },
    {
      "front": "Why is computing **'at least one'** usually easier through the complement, and what is the rule?",
      "back": "Directly summing the disjoint cases (exactly $1$, exactly $2$, …) is tedious, but 'at least one' is the complement of 'none.' For independent trials, $\\Pr(\\text{at least one})=1-\\Pr(\\text{none})=1-\\prod_{i}\\Pr(\\text{not occurring on trial }i)$.",
      "tag": "Equally likely & counting"
    },
    {
      "front": "A multiple-choice question has $5$ choices; a student answers $8$ such questions purely at random. Find $\\Pr(\\text{at least one correct})$.",
      "back": "Each question is correct with probability $1/5$, wrong with $4/5$, independently. 'At least one correct' is the complement of 'all wrong':\n$\\Pr(\\text{all wrong})=\\left(\\frac{4}{5}\\right)^{8}=\\frac{65536}{390625}\\approx 0.1678$.\n$\\Pr(\\text{at least one correct})=1-0.1678\\approx 0.8322$.",
      "tag": "Equally likely & counting"
    },
    {
      "front": "Two fair six-sided dice are rolled. Using the sample space of $36$ equally likely ordered pairs, find $\\Pr(\\text{sum}=7)$ and $\\Pr(\\text{sum}=7\\text{ or }11)$.",
      "back": "Sum $=7$: pairs $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$ — $6$ outcomes, so $\\Pr=\\frac{6}{36}=\\frac{1}{6}$.\nSum $=11$: $(5,6),(6,5)$ — $2$ outcomes. These are disjoint from 'sum $=7$', so $\\Pr(7\\text{ or }11)=\\frac{6+2}{36}=\\frac{8}{36}=\\frac{2}{9}\\approx 0.2222$.",
      "tag": "Equally likely & counting"
    },
    {
      "front": "How many distinct ways can the $11$-letter sequence in card 'MISSISSIPPI' be arranged so that it **begins with M**? (Reuse the multiset arrangement idea.)",
      "back": "Fix M in the first position; arrange the remaining $10$ letters $\\text{I}{:}4,\\ \\text{S}{:}4,\\ \\text{P}{:}2$:\n$\\frac{10!}{4!\\,4!\\,2!}=\\frac{3{,}628{,}800}{24\\cdot 24\\cdot 2}=\\frac{3{,}628{,}800}{1152}=3150$. (Equivalently $34{,}650\\times\\frac{1}{11}=3150$, since by symmetry $1$ of every $11$ arrangements starts with the single M.)",
      "tag": "Multinomial & partitions"
    }
  ]
}