Probability — NDA Mathematics

Formulas (7)

• Classical (theoretical) probability · Classical probability
  $$P(E) = \dfrac{n(E)}{n(S)} = \dfrac{\text{favourable outcomes}}{\text{total outcomes}}$$
• Axioms, range, complement, and odds · Complement rule and range
  $$0 \le P(E) \le 1, \qquad P(S)=1, \qquad P(E') = 1 - P(E)$$
• Selection probability with combinations · Selection probability (combinations)
  $$P = \dfrac{\dbinom{a}{k}\dbinom{b}{r-k}}{\dbinom{a+b}{r}}$$
• Probability with dice · Two-dice sample space
  $$n(S) = 6^2 = 36, \qquad P(\text{sum}=7) = \dfrac{6}{36} = \dfrac{1}{6}$$
• Probability with coins · Coin tosses
  $$n(S)=2^{n}, \qquad P(\text{at least one head in } n) = 1 - \left(\tfrac{1}{2}\right)^{n}$$
• Probability with arrangements · Arrangement probability (two together)
  $$P(\text{two specified together}) = \dfrac{2\,(n-1)!}{n!} = \dfrac{2}{n}$$
• Choosing numbers with a property · Counting favourable numbers
  $$P = \dfrac{\#\{x : x \text{ has the property}\}}{n}$$

Watch out for (16)

• An event is a SET of outcomes, not a single outcome
  → What is probability? (Random experiments, sample space, events)
• Write or size the sample space before you count anything
  → What is probability? (Random experiments, sample space, events)
• The classical formula needs EQUALLY LIKELY outcomes
  → Classical (theoretical) probability
• Favourable is a subset of total, so $P$ can never exceed 1
  → Classical (theoretical) probability
• Odds are not probability: $a:b$ in favour means $\dfrac{a}{a+b}$, not $\dfrac{a}{b}$
  → Axioms, range, complement, and odds
• "At least one…" almost always means use the complement
  → Axioms, range, complement, and odds
• "Drawn together / selected at random" means order does NOT matter — use $\binom{n}{r}$, not permutations
  → Selection probability with combinations
• "At least one of a type" is fastest via the complement
  → Selection probability with combinations
• Two-dice outcomes are ORDERED pairs: $(2,3)$ and $(3,2)$ are different
  → Probability with dice
• Loaded or non-standard dice: faces are NOT equally likely
  → Probability with dice
• Sequences are ordered: $HT \ne TH$
  → Probability with coins
• Biased coin: do not use $2^n$ equally-likely counting
  → Probability with coins
• "Together" = glue into a block, then multiply by the block's internal arrangements
  → Probability with arrangements
• Repeated letters divide the total by the factorial of each repeat count
  → Probability with arrangements
• "Or" on number properties needs inclusion-exclusion
  → Choosing numbers with a property
• Several numbers chosen at once: denominator is $\binom{n}{r}$, not $n$
  → Choosing numbers with a property

Formulas (4)

• The addition rule (inclusion-exclusion) · Addition rule
  $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
• "Neither" and the complement of a union · Complement of a union (De Morgan)
  $$P(A' \cap B') = 1 - P(A \cup B) = 1 - P(A) - P(B) + P(A \cap B)$$
• Mutually exclusive (disjoint) events · Addition rule for mutually exclusive events
  $$A \cap B = \varnothing \;\Rightarrow\; P(A \cup B) = P(A) + P(B)$$
• Exhaustive events (and probabilities that sum to 1) · Mutually exclusive AND exhaustive
  $$A_1 \cup \dots \cup A_n = S \;\text{ and disjoint} \;\Rightarrow\; \sum_{i} P(A_i) = 1$$

Watch out for (10)

• $A \cup B$ is INCLUSIVE "or" — it contains the overlap
  → Events as sets: union, intersection, complement
• Read "and" as intersection, "or" as union — do not swap
  → Events as sets: union, intersection, complement
• Do not forget to subtract $P(A \cap B)$
  → The addition rule (inclusion-exclusion)
• Three events need the full inclusion-exclusion, not just three single terms
  → The addition rule (inclusion-exclusion)
• "Neither" is $1 - P(A \cup B)$, not $1 - P(A) - P(B)$
  → "Neither" and the complement of a union
• De Morgan flips the operation: complement of a union is an intersection
  → "Neither" and the complement of a union
• Mutually exclusive $\ne$ independent — opposite ideas
  → Mutually exclusive (disjoint) events
• Only drop the overlap term when you are TOLD the events are mutually exclusive
  → Mutually exclusive (disjoint) events
• Exhaustive alone does not give sum $= 1$
  → Exhaustive events (and probabilities that sum to 1)
• Turn a chained ratio into one variable before summing
  → Exhaustive events (and probabilities that sum to 1)

Formulas (4)

• Independence and the multiplication rule · Multiplication rule (independent events)
  $$P(A \cap B) = P(A)\,P(B)$$
• "At least one" via the complement · At least one (independent trials)
  $$P(\text{at least one}) = 1 - \prod_{i}\big(1 - P(A_i)\big)$$
• The "problem solved by students" archetype · Problem solved by at least one solver
  $$P(\text{solved}) = 1 - \prod_{i}(1 - p_i)$$
• Finding an unknown probability using independence · Union of independent events
  $$P(A \cup B) = P(A) + P(B) - P(A)P(B) = 1 - P(A')\,P(B')$$

Watch out for (8)

• Independent $\ne$ mutually exclusive
  → Independence and the multiplication rule
• Only multiply when independence is given or physically clear
  → Independence and the multiplication rule
• "At least one" = 1 - (all fail), not the sum of individual probabilities
  → "At least one" via the complement
• Multiply the FAILURE probabilities, not the success ones, for "none"
  → "At least one" via the complement
• "Problem solved" means at least one solver — use the complement
  → The "problem solved by students" archetype
• "Exactly one solves" is a different computation
  → The "problem solved by students" archetype
• For independent events the union is NOT $P(A) + P(B)$
  → Finding an unknown probability using independence
• Use $1 - P(A')P(B')$ as a fast check
  → Finding an unknown probability using independence

Formulas (4)

• Conditional probability · Conditional probability
  $$P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}, \qquad P(B) > 0$$
• Multiplication rule & restricted sample space · Multiplication rule / restricted counting
  $$P(A \cap B) = P(B)\,P(A \mid B), \qquad P(A \mid B) = \dfrac{n(A \cap B)}{n(B)}$$
• Total probability (over a partition) · Total probability
  $$P(A) = \sum_{i=1}^{n} P(B_i)\,P(A \mid B_i)$$
• Bayes' theorem (reversing the conditional) · Bayes' theorem
  $$P(B_k \mid A) = \dfrac{P(B_k)\,P(A \mid B_k)}{\displaystyle\sum_{i} P(B_i)\,P(A \mid B_i)}$$

Watch out for (8)

• Mind which event is the condition: $P(A\mid B) \ne P(B\mid A)$ in general
  → Conditional probability
• The condition must have positive probability
  → Conditional probability
• Under a condition, the TOTAL changes to $n(B)$, not 36 (or 6)
  → Multiplication rule & restricted sample space
• The general multiplication rule needs $P(A \mid B)$, not $P(A)$
  → Multiplication rule & restricted sample space
• Weight each route by its own probability
  → Total probability (over a partition)
• The routes must be a partition
  → Total probability (over a partition)
• Do not confuse $P(B_k \mid A)$ with $P(A \mid B_k)$
  → Bayes' theorem (reversing the conditional)
• The denominator is the FULL total probability, not just $P(A \mid B_k)$
  → Bayes' theorem (reversing the conditional)

Formulas (3)

• Fréchet and Boole bounds · Bounds on intersection and union
  $$\max\big(0,\,P(A)+P(B)-1\big) \le P(A \cap B) \le \min\big(P(A),\,P(B)\big)$$
• Minimum and maximum of combined probabilities · Linking identity
  $$P(A) + P(B) = P(A \cup B) + P(A \cap B)$$
• Identity-statement traps ("which are correct?") · Exactly one vs union
  $$P(\text{exactly one}) = P(A) + P(B) - 2P(A \cap B), \qquad P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

Watch out for (6)

• The intersection floor $P(A)+P(B)-1$ only bites when the sum exceeds 1
  → Fréchet and Boole bounds
• The union floor is $\max(P(A),P(B))$, not $P(A)+P(B)$
  → Fréchet and Boole bounds
• Minimum union = max of the two probabilities, maximum union = their sum (capped at 1)
  → Minimum and maximum of combined probabilities
• Convert via $P(A)+P(B) = P(A\cup B) + P(A\cap B)$
  → Minimum and maximum of combined probabilities
• "Exactly one" subtracts the overlap TWICE
  → Identity-statement traps ("which are correct?")
• $P(A \cap \bar{B}) = P(A) - P(A \cap B)$, not $P(A) - P(B)$
  → Identity-statement traps ("which are correct?")