Exam P · practice
Exam P — Conditional Probability & Bayes Practice Flashcards
Thirty-three original SOA/CAS Exam P-style multiple-choice problems on conditional probability, the multiplication and chain rules, independence, the law of total probability, Bayes' theorem, diagnostic tests, and conditional independence, each with a full worked solution.
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More Exam P decks:
Common Traps Conditional Probability & Bayes Continuous Distributions Continuous Distributions Practice Covariance, Sums & CLT Covariance, Sums & CLT Practice
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- Conditional definitionAn insurer classifies its policyholders by two events: $A$ = the policyholder owns a home, $B$ = the policyholder owns a vehicle. You are given $\Pr(A)=0.55$, $\Pr(B)=0.70$, and $\Pr(A\cup B)=0.85$. Calculate $\Pr(A\mid B)$. (A) $0.40$ (B) $0.57$ (C) $0.65$ (D) $0.73$ (E) $0.86$**Answer: (B).** First recover the intersection from inclusion-exclusion: $\Pr(A\cap B)=\Pr(A)+\Pr(B)-\Pr(A\cup B)=0.55+0.70-0.85=0.40$. Then apply the definition of conditional probability: $\Pr(A\mid B)=\dfrac{\Pr(A\cap B)}{\Pr(B)}=\dfrac{0.40}{0.70}\approx 0.5714$. Distractor (A) is the raw intersection $0.40$; (C) divides by $\Pr(A)$ instead of $\Pr(B)$ ($0.40/0.55\approx0.727$ is (D)); (E) is $\Pr(A\cup B)$.
- Conditional definitionYou are given $\Pr(A\mid B)=0.30$, $\Pr(B\mid A)=0.45$, and $\Pr(A)=0.20$. Calculate $\Pr(B)$. (A) $0.06$ (B) $0.09$ (C) $0.13$ (D) $0.30$ (E) $0.45$**Answer: (D).** The joint probability can be written two ways: $\Pr(A\cap B)=\Pr(B\mid A)\,\Pr(A)=0.45\times0.20=0.09$. Also $\Pr(A\cap B)=\Pr(A\mid B)\,\Pr(B)$, so $\Pr(B)=\dfrac{\Pr(A\cap B)}{\Pr(A\mid B)}=\dfrac{0.09}{0.30}=0.30$. Distractor (B) is the joint $0.09$; (A) is $\Pr(A\mid B)\Pr(A)=0.06$ (wrong conditioning); (E) just echoes $\Pr(B\mid A)$.
- Conditional definitionA loss random variable $X$ has the property that $\Pr(X>10)=0.40$ and $\Pr(X>25)=0.10$. Given that a loss exceeds $10$, calculate the probability that it also exceeds $25$. (A) $0.10$ (B) $0.25$ (C) $0.30$ (D) $0.40$ (E) $0.67$**Answer: (B).** Because $\{X>25\}\subset\{X>10\}$, the intersection is just $\{X>25\}$: $\Pr(X>25\mid X>10)=\dfrac{\Pr(X>25\cap X>10)}{\Pr(X>10)}=\dfrac{\Pr(X>25)}{\Pr(X>10)}=\dfrac{0.10}{0.40}=0.25$. Distractor (A) forgets to condition (just reports $\Pr(X>25)$); (C) subtracts $0.40-0.10$; (E) inverts the ratio to $0.40/0.10$ truncated.
- Multiplication ruleAn urn contains 7 red and 5 blue balls. Three balls are drawn without replacement. Calculate the probability that all three are red. (A) $0.114$ (B) $0.159$ (C) $0.197$ (D) $0.292$ (E) $0.583$**Answer: (B).** Apply the chain (multiplication) rule for dependent draws: $\Pr(R_1\cap R_2\cap R_3)=\dfrac{7}{12}\cdot\dfrac{6}{11}\cdot\dfrac{5}{10}=\dfrac{210}{1320}\approx 0.1591$. Equivalently $\dfrac{\binom{7}{3}}{\binom{12}{3}}=\dfrac{35}{220}\approx0.1591$. Distractor (E) $=7/12$ counts only the first draw; (C) $\approx(7/12)^3=0.198$ treats the draws as **with replacement**; (A) and (D) come from miscounting the urn (wrong number of balls in the denominator).
- Multiplication ruleA deck of 52 cards is shuffled and cards are dealt one at a time without replacement. Calculate the probability that the first three cards are all hearts. (A) $0.0129$ (B) $0.0156$ (C) $0.0173$ (D) $0.0588$ (E) $0.2500$**Answer: (A).** There are 13 hearts. By the chain rule: $\Pr=\dfrac{13}{52}\cdot\dfrac{12}{51}\cdot\dfrac{11}{50}=\dfrac{1716}{132600}\approx 0.01294$. Distractor (B) is $(13/52)^3=(1/4)^3=0.015625$ (with-replacement error); (E) is the single-card $13/52=1/4$; (D) is $\frac{12}{51}\cdot\frac{11}{50}$ style miscount.
- IndependenceAn actuary models three independent system components $C_1,C_2,C_3$ each operating with probability $0.92$. The system works only if all three components work. Calculate the probability the system works. (A) $0.760$ (B) $0.779$ (C) $0.847$ (D) $0.920$ (E) $0.998$**Answer: (B).** For independent components, joint operation is the product: $\Pr(\text{all work})=0.92^3=0.778688\approx 0.779$. Distractor (C) is $0.92^2\approx0.846$ (only two components); (E) is $1-(1-0.92)^3=0.999488$ rounded, the at-least-one formula (wrong series-vs-parallel); (A) is $3(0.92)-2$ style error.
- IndependenceEvents $A$ and $B$ are independent with $\Pr(A)=0.4$ and $\Pr(B)=0.5$. Calculate $\Pr(A\cup B)$. (A) $0.20$ (B) $0.50$ (C) $0.70$ (D) $0.80$ (E) $0.90$**Answer: (C).** Independence gives $\Pr(A\cap B)=\Pr(A)\Pr(B)=0.4\times0.5=0.20$. Then $\Pr(A\cup B)=\Pr(A)+\Pr(B)-\Pr(A\cap B)=0.4+0.5-0.20=0.70$. Distractor (E) is $0.4+0.5$ (forgets to subtract the intersection); (A) is the intersection $0.20$; (D) wrongly treats them as mutually exclusive complements.
- IndependenceA device has two independent backup batteries. Each battery fails during a mission with probability $0.15$. The device fails only if both batteries fail. Calculate the probability the device fails. (A) $0.0225$ (B) $0.150$ (C) $0.255$ (D) $0.300$ (E) $0.7225$**Answer: (A).** Independent failures: $\Pr(\text{both fail})=0.15\times0.15=0.0225$. Distractor (D) is $2(0.15)=0.30$ (adds, ignoring independence/overlap); (C) is $1-0.15^2$ inverted reasoning; (E) is $0.85^2$, the probability both survive.