Exam P — Set Theory & Combinatorics Practice Flashcards

Thirty-three SOA-style multiple-choice problems covering set operations, De Morgan, Kolmogorov axioms, inclusion-exclusion, equally likely counting, permutations, combinations, multinomial coefficients, and the binomial theorem.

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Inclusion-exclusion
An auto insurer classifies each policyholder by whether they have filed a comprehensive claim ($A$) and whether they have filed a collision claim ($B$). For a randomly chosen policyholder, $\Pr(A)=0.36$, $\Pr(B)=0.28$, and $\Pr(A\cap B)=0.12$. Calculate the probability that the policyholder has filed at least one of the two types of claim. (A) $0.40$ (B) $0.52$ (C) $0.64$ (D) $0.76$ (E) $0.88$
**Answer: (B).** Use the addition rule (inclusion-exclusion for two events): $\Pr(A\cup B)=\Pr(A)+\Pr(B)-\Pr(A\cap B)$. Here $\Pr(A\cup B)=0.36+0.28-0.12=0.52$. (Choice (C), $0.64$, drops the subtraction of the intersection.)
Set operations
For events $A$ and $B$ in a sample space $S$, you are given $\Pr(A)=0.55$, $\Pr(B)=0.45$, and $\Pr(A\cup B)=0.80$. Calculate $\Pr(A^{c}\cap B^{c})$. (A) $0.00$ (B) $0.20$ (C) $0.25$ (D) $0.35$ (E) $0.45$
**Answer: (B).** By De Morgan's law, $A^{c}\cap B^{c}=(A\cup B)^{c}$, so $\Pr(A^{c}\cap B^{c})=1-\Pr(A\cup B)=1-0.80=0.20$. (Choice (C), $0.25$, comes from $1-\Pr(A)-\Pr(B)+\Pr(A\cup B)$ or other miscombinations of the marginals.)
Inclusion-exclusion
A health plan covers dental ($D$), vision ($V$), and prescription ($R$) benefits as optional riders. Among enrollees, $\Pr(D)=\Pr(V)=\Pr(R)=0.40$, each pairwise intersection has probability $0.15$, and $\Pr(D\cap V\cap R)=0.05$. Calculate the probability that a randomly chosen enrollee has at least one rider. (A) $0.65$ (B) $0.70$ (C) $0.75$ (D) $0.80$ (E) $0.95$
**Answer: (D).** Three-event inclusion-exclusion: $\Pr(D\cup V\cup R)=\sum\Pr(\text{single})-\sum\Pr(\text{pairwise})+\Pr(\text{triple})$. So $\Pr(D\cup V\cup R)=3(0.40)-3(0.15)+0.05=1.20-0.45+0.05=0.80$. (Choice (E), $0.95$, forgets to subtract the pairwise terms; choice (C), $0.75$, drops the triple add-back.)
Inclusion-exclusion
A health plan offers three optional riders with $\Pr(D)=\Pr(V)=\Pr(R)=0.40$, each pairwise intersection $0.15$, and triple intersection $\Pr(D\cap V\cap R)=0.05$. Calculate the probability that an enrollee has **exactly one** rider. (A) $0.30$ (B) $0.40$ (C) $0.45$ (D) $0.50$ (E) $0.60$
**Answer: (C).** The probability of "exactly one" for three events is $\sum\Pr(\text{single})-2\sum\Pr(\text{pairwise})+3\Pr(\text{triple})$. So the value is $3(0.40)-2(3)(0.15)+3(0.05)=1.20-0.90+0.15=0.45$. (Choice (A), $0.30$, uses the "at least one" complement logic incorrectly; choice (E), $0.60$, subtracts the pairwise terms only once.)
Axioms
An actuary models two events with $\Pr(A)=0.6$ and $\Pr(B)=0.5$. The events are **not** necessarily disjoint or independent. Which of the following is a possible value of $\Pr(A\cap B)$? (A) $0.00$ (B) $0.05$ (C) $0.45$ (D) $0.65$ (E) $0.80$
**Answer: (C).** The intersection probability must satisfy the bounds $\max(0,\ \Pr(A)+\Pr(B)-1)\le \Pr(A\cap B)\le \min(\Pr(A),\Pr(B))$. Here $\Pr(A)+\Pr(B)-1=0.6+0.5-1=0.1$ and $\min(0.6,0.5)=0.5$, so $0.1\le\Pr(A\cap B)\le 0.5$. Only $0.45$ lies in $[0.1,0.5]$. Choices (A) and (B) fall below the lower bound (the events overlap too much to be that small), and (D), (E) exceed $\min(\Pr(A),\Pr(B))$.
Axioms
A six-sided die is loaded so that the probability of each face is proportional to the number of pips on that face. That is, $\Pr(\text{face } k)=ck$ for $k=1,2,\ldots,6$. Calculate $\Pr(\text{the roll is even})$. (A) $\dfrac{6}{21}$ (B) $\dfrac{9}{21}$ (C) $\dfrac{10}{21}$ (D) $\dfrac{11}{21}$ (E) $\dfrac{12}{21}$
**Answer: (E).** Normalize: $\sum_{k=1}^{6}ck=c(1+2+3+4+5+6)=21c=1$, so $c=\tfrac{1}{21}$. The even faces are $2,4,6$, so $\Pr(\text{even})=c(2+4+6)=\dfrac{12}{21}=\dfrac{4}{7}$. (Choice (B), $\tfrac{9}{21}$, sums the odd faces $1+3+5$ by mistake; choice (C), $\tfrac{10}{21}$, omits face $2$ from the even sum.)
Set operations
In a probability model, $A\subseteq B$. You are also told $\Pr(A)=0.30$ and $\Pr(B)=0.45$. Calculate $\Pr(B\cap A^{c})$, the probability that $B$ occurs but $A$ does not. (A) $0.00$ (B) $0.15$ (C) $0.30$ (D) $0.45$ (E) $0.55$
**Answer: (B).** Because $A\subseteq B$, we can write $B$ as the disjoint union $B=A\cup(B\cap A^{c})$. Then $\Pr(B)=\Pr(A)+\Pr(B\cap A^{c})$, so $\Pr(B\cap A^{c})=0.45-0.30=0.15$. (Choice (D) just restates $\Pr(B)$; choice (E), $0.55$, computes $1-\Pr(A)-\Pr(B\cap A^c)$-style nonsense.)
Set operations
A survey of policyholders finds that $70\%$ own a home, $60\%$ own a car, and $50\%$ own both. Calculate the percentage who own a home but **not** a car. (A) $10\%$ (B) $20\%$ (C) $30\%$ (D) $40\%$ (E) $50\%$
**Answer: (B).** Let $H$ = owns home, $C$ = owns car. We want $\Pr(H\cap C^{c})$. Decompose $H$ into the disjoint pieces $H\cap C$ and $H\cap C^{c}$: $\Pr(H\cap C^{c})=\Pr(H)-\Pr(H\cap C)=0.70-0.50=0.20$. So $20\%$. (Choice (A), $10\%$, mistakenly uses $\Pr(C)-\Pr(H\cap C)$; choice (C), $30\%$, uses $1-\Pr(H)$.)