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Exam P — Discrete Distributions Practice Flashcards

Thirty-three original SOA/CAS Exam P-style multiple-choice problems spanning pmf/cdf mechanics, the binomial family, waiting-time distributions, hypergeometric sampling, Poisson processes, and moment/MGF identification.

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Common Traps Conditional Probability & Bayes Conditional Probability & Bayes Practice Continuous Distributions Continuous Distributions Practice Covariance, Sums & CLT

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Browse all 33 problems as a list
  1. pmf/cdf
    A discrete random variable $X$ has probability mass function $p(x)=c\left(\dfrac{2}{3}\right)^{x}$ for $x=1,2,3,\dots$, and $p(x)=0$ otherwise. Determine $\Pr(X \geq 2)$. (A) $0.111$ (B) $0.333$ (C) $0.444$ (D) $0.500$ (E) $0.667$
    **Answer: (E).** First find $c$ from total mass one. The support is a geometric series starting at $x=1$: $\sum_{x=1}^{\infty}c\left(\tfrac{2}{3}\right)^{x}=c\cdot\dfrac{2/3}{1-2/3}=c\cdot 2=1$, so $c=\tfrac{1}{2}$. Then $p(1)=\tfrac{1}{2}\cdot\tfrac{2}{3}=\tfrac{1}{3}$, so $\Pr(X \geq 2)=1-\Pr(X=1)=1-\tfrac{1}{3}=\tfrac{2}{3}\approx 0.667$. (Choice (B) is $p(1)$ itself; (A) mistakes $\Pr(X=2)$.)
  2. pmf/cdf
    A discrete random variable $N$ has cumulative distribution function $F(n)=\begin{cases}0 & n<0\\ 0.2 & 0 \leq n<1\\ 0.5 & 1 \leq n<2\\ 0.9 & 2 \leq n<3\\ 1 & n \geq 3\end{cases}$ Calculate $\Pr(1 \leq N \leq 2)$. (A) $0.30$ (B) $0.40$ (C) $0.50$ (D) $0.70$ (E) $0.90$
    **Answer: (D).** The pmf is read from the jumps: $\Pr(N=1)=0.5-0.2=0.3$ and $\Pr(N=2)=0.9-0.5=0.4$. $\Pr(1 \leq N \leq 2)=\Pr(N=1)+\Pr(N=2)=0.3+0.4=0.7$. Equivalently $F(2)-F(1^{-})=0.9-0.2=0.7$. (Choice (B) uses $F(2)-F(1)=0.9-0.5=0.4$, dropping the mass at $N=1$.)
  3. pmf/cdf
    A non-negative integer-valued random variable $X$ has survival function $\Pr(X>k)=\left(\dfrac{3}{4}\right)^{k+1}$ for $k=0,1,2,\dots$. Use the tail-sum formula to calculate $E[X]$. (A) $1.33$ (B) $2.25$ (C) $3.00$ (D) $3.75$ (E) $4.00$
    **Answer: (C).** The tail-sum formula is $E[X]=\sum_{k=0}^{\infty}\Pr(X>k)$. $E[X]=\sum_{k=0}^{\infty}\left(\tfrac{3}{4}\right)^{k+1}=\tfrac{3}{4}\sum_{k=0}^{\infty}\left(\tfrac{3}{4}\right)^{k}=\tfrac{3}{4}\cdot\dfrac{1}{1-3/4}=\tfrac{3}{4}\cdot 4=3$. (Choice (D) sums $\sum_{k=1}^{\infty}\Pr(X>k)$ incorrectly; (E) forgets the leading $\tfrac{3}{4}$ factor.)
  4. Binomial Family
    An integer $X$ is drawn from a discrete uniform distribution on $\{1,2,\dots,12\}$. Calculate $\operatorname{Var}(X)$. (A) $6.5$ (B) $11.92$ (C) $12.00$ (D) $14.33$ (E) $42.25$
    **Answer: (B).** For a discrete uniform on $\{1,\dots,n\}$, $\operatorname{Var}(X)=\dfrac{n^{2}-1}{12}$. With $n=12$: $\operatorname{Var}(X)=\dfrac{144-1}{12}=\dfrac{143}{12}\approx 11.92$. (Choice (A) is the mean $\tfrac{n+1}{2}=6.5$; (C) uses $\tfrac{n^2}{12}$ forgetting the $-1$; (E) is the mean squared.)
  5. Binomial Family
    An insurer issues $15$ identical policies, each of which independently files a claim during the year with probability $0.12$. Let $X$ be the number of policies that file a claim. Calculate $\Pr(X=2)$. (A) $0.144$ (B) $0.211$ (C) $0.287$ (D) $0.296$ (E) $0.339$
    **Answer: (C).** $X\sim\text{Binomial}(15,0.12)$. $\Pr(X=2)=\binom{15}{2}(0.12)^{2}(0.88)^{13}$. $\binom{15}{2}=105$, $(0.12)^{2}=0.0144$, and $(0.88)^{13}\approx 0.18979$. $\Pr(X=2)=105\cdot 0.0144\cdot 0.18979\approx 0.287$. (Choice (E) uses the wrong failure exponent $(0.88)^{12}$ instead of $(0.88)^{13}$; the other distractors are nearby binomial masses from off-by-one errors in $x$.)
  6. Binomial Family
    A multiple-choice quiz has $20$ questions, each with $5$ choices. A student guesses every answer independently. Let $X$ be the number correct. Calculate $E[X]$ and $\operatorname{Var}(X)$. (A) $E[X]=4$, $\operatorname{Var}(X)=3.2$ (B) $E[X]=4$, $\operatorname{Var}(X)=4.0$ (C) $E[X]=4$, $\operatorname{Var}(X)=1.6$ (D) $E[X]=5$, $\operatorname{Var}(X)=3.2$ (E) $E[X]=5$, $\operatorname{Var}(X)=4.0$
    **Answer: (A).** $X\sim\text{Binomial}(n,p)$ with $n=20$, $p=\tfrac{1}{5}=0.2$. Mean: $E[X]=np=20(0.2)=4$. Variance: $\operatorname{Var}(X)=npq=20(0.2)(0.8)=3.2$. (Choice (B) uses $\operatorname{Var}=np$ forgetting the $q$ factor; (C) uses $npq$ with $p$ and $q$ swapped incorrectly as $np^2$; (D)–(E) misread $n/p$.)
  7. Binomial Family
    In a portfolio, each of $6$ independent risks suffers a loss with probability $0.4$. Calculate the probability that at least $5$ of the risks suffer a loss. (A) $0.041$ (B) $0.179$ (C) $0.233$ (D) $0.456$ (E) $0.767$
    **Answer: (A).** $X\sim\text{Binomial}(6,0.4)$, want $\Pr(X \geq 5)=\Pr(X=5)+\Pr(X=6)$. $\Pr(X=5)=\binom{6}{5}(0.4)^{5}(0.6)=6(0.01024)(0.6)=0.0368640$. $\Pr(X=6)=(0.4)^{6}=0.0040960$. $\Pr(X \geq 5)=0.0368640+0.0040960=0.0409600\approx 0.041$. (Choice (B) is $\Pr(X=4)$; (C) is $\Pr(X \geq 4)$.)
  8. Binomial Family
    A biased coin lands heads with probability $0.45$. It is tossed $7$ times independently. Given that at least one head appears, calculate the probability that exactly one head appears. (A) $0.045$ (B) $0.089$ (C) $0.102$ (D) $0.152$ (E) $0.305$
    **Answer: (B).** Let $X\sim\text{Binomial}(7,0.45)$. We want $\Pr(X=1 \mid X \geq 1)=\dfrac{\Pr(X=1)}{\Pr(X \geq 1)}$. $\Pr(X=1)=\binom{7}{1}(0.45)(0.55)^{6}=7(0.45)(0.027681)=0.087195$. $\Pr(X \geq 1)=1-(0.55)^{7}=1-0.015224=0.984776$. $\Pr(X=1 \mid X \geq 1)=\dfrac{0.087195}{0.984776}\approx 0.0885$, i.e. $\approx 0.089$. (Choice (D) reports the two-head conditional probability; (C) divides by $\Pr(X \geq 1)$ using $\binom{7}{1}$ but with $p$ and $q$ exponents swapped.)