{
  "deckName": "Exam P — Discrete Distributions",
  "examCode": "Exam P",
  "cards": [
    {
      "front": "What two conditions must a function $p(x)$ satisfy to be a valid probability mass function (pmf)?",
      "back": "(1) Non-negativity: $p(x) \\geq 0$ for every $x$ in the support.\n(2) Total mass one: $\\sum_{x} p(x) = 1$, summed over all support points.\nValues outside the support have $p(x)=0$.",
      "tag": "pmf/cdf"
    },
    {
      "front": "For a discrete random variable, how is the cumulative distribution function $F(x)$ built from the pmf, and what does $\\Pr(a < X \\leq b)$ equal?",
      "back": "The cdf accumulates mass: $F(x)=\\Pr(X \\leq x)=\\sum_{k \\leq x} p(k)$.\nIt is a right-continuous non-decreasing step function with $F(-\\infty)=0$ and $F(\\infty)=1$.\nFor a discrete variable, $\\Pr(a < X \\leq b)=F(b)-F(a)$.",
      "tag": "pmf/cdf"
    },
    {
      "front": "A discrete variable has cdf $F$. Express the point mass $\\Pr(X=x)$ in terms of $F$, and the survival probability for an integer-valued variable.",
      "back": "The jump in the cdf gives the point mass: $\\Pr(X=x)=F(x)-F(x^{-})$, the size of the step at $x$.\nThe survival function is $S(x)=\\Pr(X>x)=1-F(x)$, so for integer support $\\Pr(X \\geq x)=1-F(x-1)$.",
      "tag": "pmf/cdf"
    },
    {
      "front": "State the tail-sum (survival) formula for the expectation of a non-negative integer-valued random variable $X$.",
      "back": "$E[X]=\\sum_{k=1}^{\\infty}\\Pr(X \\geq k)=\\sum_{k=0}^{\\infty}\\Pr(X>k)$.\nThis sums the survival function and is often faster than $\\sum_{k} k\\,p(k)$ — it is the discrete analogue of $E[X]=\\int_{0}^{\\infty}S(x)\\,dx$.",
      "tag": "pmf/cdf"
    },
    {
      "front": "Contrast binomial, negative binomial/geometric, and hypergeometric in terms of what is fixed and what is counted.",
      "back": "**Binomial:** fixed number of independent trials $n$; count the successes; replacement (constant $p$).\n**Geometric / negative binomial:** fixed number of successes ($1$ or $r$); count the trials (or failures) until you reach them.\n**Hypergeometric:** fixed sample size $n$ drawn **without replacement**; count successes from a finite population — trials are dependent.",
      "tag": "pmf/cdf"
    },
    {
      "front": "Give the pmf, mean, and variance of a discrete uniform distribution on $\\{1,2,\\dots,n\\}$.",
      "back": "pmf: $p(x)=\\dfrac{1}{n}$ for $x=1,\\dots,n$.\nMean: $E[X]=\\dfrac{n+1}{2}$.\nVariance: $\\mathrm{Var}(X)=\\dfrac{n^{2}-1}{12}$.\nEach outcome is equally likely (e.g. a fair die has $n=6$).",
      "tag": "Binomial Family"
    },
    {
      "front": "A fair six-sided die is rolled once. Find $E[X]$ and $\\mathrm{Var}(X)$ where $X$ is the face shown.",
      "back": "This is discrete uniform on $\\{1,\\dots,6\\}$, so $n=6$.\nMean: $E[X]=\\dfrac{n+1}{2}=\\dfrac{7}{2}=3.5$.\nVariance: $\\mathrm{Var}(X)=\\dfrac{n^{2}-1}{12}=\\dfrac{35}{12}\\approx 2.917$.",
      "tag": "Binomial Family"
    },
    {
      "front": "Define a Bernoulli$(p)$ random variable and give its mean, variance, and MGF.",
      "back": "It takes value $1$ ('success') with probability $p$ and $0$ ('failure') with probability $q=1-p$.\nMean: $E[X]=p$.\nVariance: $\\mathrm{Var}(X)=pq=p(1-p)$.\nMGF: $M_X(t)=q+pe^{t}$.",
      "tag": "Binomial Family"
    },
    {
      "front": "For a Bernoulli$(p)$ variable, what value of $p$ maximizes the variance, and what is that maximum?",
      "back": "The variance $\\mathrm{Var}(X)=p(1-p)$ is a downward parabola in $p$, maximized at $p=\\tfrac{1}{2}$.\nMaximum value: $\\mathrm{Var}(X)=\\tfrac{1}{2}\\cdot\\tfrac{1}{2}=0.25$.",
      "tag": "Binomial Family"
    },
    {
      "front": "Give the pmf, mean, variance, and MGF of a Binomial$(n,p)$ random variable.",
      "back": "pmf: $p(x)=\\binom{n}{x}p^{x}q^{\\,n-x}$ for $x=0,1,\\dots,n$, where $q=1-p$.\nMean: $E[X]=np$.\nVariance: $\\mathrm{Var}(X)=npq$.\nMGF: $M_X(t)=(q+pe^{t})^{n}$.\nIt counts successes in $n$ independent identical trials.",
      "tag": "Binomial Family"
    },
    {
      "front": "A salesperson makes $10$ independent calls, each closing with probability $0.3$. Find the probability of exactly $4$ sales.",
      "back": "Let $X\\sim\\text{Binomial}(10,0.3)$.\n$\\Pr(X=4)=\\binom{10}{4}(0.3)^{4}(0.7)^{6}$.\n$\\binom{10}{4}=210$, so $\\Pr(X=4)=210\\,(0.0081)(0.117649)\\approx 0.2001$.",
      "tag": "Binomial Family"
    },
    {
      "front": "A binomial experiment has $n=8$ trials with success probability $p=0.25$. What is the probability of at least one success?",
      "back": "Use the complement of 'no successes':\n$\\Pr(X \\geq 1)=1-\\Pr(X=0)=1-(0.75)^{8}$.\n$(0.75)^{8}\\approx 0.1001$, so $\\Pr(X \\geq 1)\\approx 0.8999$.",
      "tag": "Binomial Family"
    },
    {
      "front": "For $X\\sim\\text{Binomial}(5,0.2)$, compute $\\Pr(X \\geq 2)$.",
      "back": "Subtract the $0$ and $1$ terms from one:\n$\\Pr(X \\geq 2)=1-\\Pr(X=0)-\\Pr(X=1)$.\n$\\Pr(X=0)=(0.8)^{5}=0.32768$; $\\Pr(X=1)=\\binom{5}{1}(0.2)(0.8)^{4}=5(0.2)(0.4096)=0.4096$.\n$\\Pr(X \\geq 2)=1-0.32768-0.4096\\approx 0.2627$.",
      "tag": "Binomial Family"
    },
    {
      "front": "Geometric distribution, convention A (number of trials until the first success). Give the pmf, mean, variance, and MGF.",
      "back": "Support $\\{1,2,3,\\dots\\}$ with $X=$ trial of first success.\npmf: $p(x)=q^{x-1}p$, where $q=1-p$.\nMean: $E[X]=\\dfrac{1}{p}$.\nVariance: $\\mathrm{Var}(X)=\\dfrac{q}{p^{2}}$.\nMGF: $M_X(t)=\\dfrac{pe^{t}}{1-qe^{t}}$ for $t<-\\ln q$.",
      "tag": "Waiting-Time"
    },
    {
      "front": "Geometric distribution, convention B (number of failures before the first success). Give the pmf, mean, and variance, and contrast with convention A.",
      "back": "Support $\\{0,1,2,\\dots\\}$ with $Y=$ failures before the first success.\npmf: $p(y)=q^{y}p$.\nMean: $E[Y]=\\dfrac{q}{p}$.\nVariance: $\\mathrm{Var}(Y)=\\dfrac{q}{p^{2}}$ (same as convention A).\nNote $Y=X-1$, so the means differ by exactly $1$: $\\tfrac{1}{p}$ vs $\\tfrac{q}{p}$. Always read the problem to pick the support.",
      "tag": "Waiting-Time"
    },
    {
      "front": "A fair die is rolled until the first six appears. Let $X$ be the number of rolls (convention A). Find $E[X]$ and $\\Pr(X=4)$.",
      "back": "Here $p=\\tfrac{1}{6}$, $q=\\tfrac{5}{6}$.\nMean: $E[X]=\\dfrac{1}{p}=6$.\n$\\Pr(X=4)=q^{3}p=\\left(\\tfrac{5}{6}\\right)^{3}\\tfrac{1}{6}=\\dfrac{125}{1296}\\approx 0.0965$.",
      "tag": "Waiting-Time"
    },
    {
      "front": "For a geometric variable (convention A) with $p=0.2$, give the variance and the survival probability $\\Pr(X>5)$.",
      "back": "Variance: $\\mathrm{Var}(X)=\\dfrac{q}{p^{2}}=\\dfrac{0.8}{0.04}=20$.\nSurvival: $\\Pr(X>k)=q^{k}$, so $\\Pr(X>5)=(0.8)^{5}=0.32768$.",
      "tag": "Waiting-Time"
    },
    {
      "front": "State the memorylessness property of the geometric distribution and give the formula.",
      "back": "Past failures do not change future waiting time: for $s,t \\geq 0$,\n$\\Pr(X>s+t \\mid X>s)=\\Pr(X>t)$.\nThe geometric is the only discrete distribution with this property (the exponential is its continuous analogue). It follows from $\\Pr(X>k)=q^{k}$.",
      "tag": "Waiting-Time"
    },
    {
      "front": "A component's lifetime in whole weeks is geometric (convention A) with $p=0.25$. Given it has already survived past week $2$, find the probability it survives past week $5$.",
      "back": "By memorylessness, $\\Pr(X>5 \\mid X>2)=\\Pr(X>3)$.\nUsing $\\Pr(X>k)=q^{k}$ with $q=0.75$:\n$\\Pr(X>3)=(0.75)^{3}=0.421875\\approx 0.4219$.",
      "tag": "Waiting-Time"
    },
    {
      "front": "Negative Binomial$(r,p)$, 'trials until the $r$-th success' form. Give the pmf, mean, variance, and MGF.",
      "back": "Support $\\{r,r+1,\\dots\\}$ with $X=$ trial of the $r$-th success.\npmf: $p(x)=\\binom{x-1}{r-1}p^{r}q^{\\,x-r}$, $q=1-p$.\nMean: $E[X]=\\dfrac{r}{p}$.\nVariance: $\\mathrm{Var}(X)=\\dfrac{rq}{p^{2}}$.\nMGF: $M_X(t)=\\left(\\dfrac{pe^{t}}{1-qe^{t}}\\right)^{r}$.\nIt reduces to the geometric when $r=1$.",
      "tag": "Waiting-Time"
    },
    {
      "front": "Negative Binomial, 'number of failures' form. For $W=$ failures before the $r$-th success, give the pmf, mean, and variance.",
      "back": "Support $\\{0,1,2,\\dots\\}$.\npmf: $p(w)=\\binom{w+r-1}{w}p^{r}q^{w}$.\nMean: $E[W]=\\dfrac{rq}{p}$.\nVariance: $\\mathrm{Var}(W)=\\dfrac{rq}{p^{2}}$ (same as the trials form).\nNote $W=X-r$, so the means differ by $r$.",
      "tag": "Waiting-Time"
    },
    {
      "front": "Independent trials succeed with probability $0.4$. Let $X$ be the trial on which the $3$rd success occurs. Find $\\Pr(X=5)$.",
      "back": "Negative binomial, trials form, $r=3$, $p=0.4$, $q=0.6$.\n$\\Pr(X=5)=\\binom{5-1}{3-1}p^{3}q^{2}=\\binom{4}{2}(0.4)^{3}(0.6)^{2}$.\n$\\binom{4}{2}=6$, so $\\Pr(X=5)=6\\,(0.064)(0.36)=0.13824\\approx 0.1382$.",
      "tag": "Waiting-Time"
    },
    {
      "front": "For the negative binomial (trials form) with $r=3$ and $p=0.4$, compute the mean and variance.",
      "back": "Mean: $E[X]=\\dfrac{r}{p}=\\dfrac{3}{0.4}=7.5$.\nVariance: $\\mathrm{Var}(X)=\\dfrac{rq}{p^{2}}=\\dfrac{3(0.6)}{0.16}=\\dfrac{1.8}{0.16}=11.25$.",
      "tag": "Waiting-Time"
    },
    {
      "front": "For $W=$ failures before the $2$nd success with $p=0.3$ (failures form), find $\\Pr(W=3)$.",
      "back": "$\\Pr(W=3)=\\binom{w+r-1}{w}p^{r}q^{w}=\\binom{3+2-1}{3}(0.3)^{2}(0.7)^{3}$.\n$\\binom{4}{3}=4$, so $\\Pr(W=3)=4\\,(0.09)(0.343)=0.12348\\approx 0.1235$.",
      "tag": "Waiting-Time"
    },
    {
      "front": "Give the pmf, mean, and variance of a Hypergeometric$(N,K,n)$ distribution.",
      "back": "Draw $n$ items without replacement from $N$ items of which $K$ are 'successes'; $X=$ successes drawn.\npmf: $p(x)=\\dfrac{\\binom{K}{x}\\binom{N-K}{\\,n-x}}{\\binom{N}{n}}$.\nMean: $E[X]=n\\dfrac{K}{N}$.\nVariance: $\\mathrm{Var}(X)=n\\dfrac{K}{N}\\left(1-\\dfrac{K}{N}\\right)\\dfrac{N-n}{N-1}$.\nThe last factor $\\tfrac{N-n}{N-1}$ is the finite-population correction.",
      "tag": "Hypergeometric"
    },
    {
      "front": "When should you use the hypergeometric distribution instead of the binomial?",
      "back": "Use the hypergeometric for sampling **without replacement** from a finite population (the trials are dependent because each draw changes the remaining composition).\nUse the binomial for independent trials with constant success probability — i.e. sampling **with replacement**, or drawing from a population so large the depletion is negligible.",
      "tag": "Hypergeometric"
    },
    {
      "front": "An urn has $20$ balls, $8$ of which are red. Five balls are drawn without replacement. Find the probability that exactly $2$ are red.",
      "back": "Hypergeometric with $N=20$, $K=8$, $n=5$.\n$\\Pr(X=2)=\\dfrac{\\binom{8}{2}\\binom{12}{3}}{\\binom{20}{5}}=\\dfrac{28\\cdot 220}{15504}=\\dfrac{6160}{15504}\\approx 0.3973$.",
      "tag": "Hypergeometric"
    },
    {
      "front": "A committee of $4$ is chosen from $7$ women and $5$ men. Find the probability of exactly $2$ women.",
      "back": "Hypergeometric: $N=12$, $K=7$ (women), $n=4$.\n$\\Pr(X=2)=\\dfrac{\\binom{7}{2}\\binom{5}{2}}{\\binom{12}{4}}=\\dfrac{21\\cdot 10}{495}=\\dfrac{210}{495}\\approx 0.4242$.",
      "tag": "Hypergeometric"
    },
    {
      "front": "A batch of $50$ parts contains $5$ defectives. Four parts are inspected without replacement. Find the probability that none is defective.",
      "back": "Hypergeometric: $N=50$, $K=5$, $n=4$.\n$\\Pr(X=0)=\\dfrac{\\binom{5}{0}\\binom{45}{4}}{\\binom{50}{4}}=\\dfrac{1\\cdot 148995}{230300}\\approx 0.6470$.",
      "tag": "Hypergeometric"
    },
    {
      "front": "Compute the mean and variance for drawing $n=5$ from $N=20$ with $K=8$ successes (hypergeometric).",
      "back": "Mean: $E[X]=n\\dfrac{K}{N}=5\\cdot\\dfrac{8}{20}=2$.\nVariance: $\\mathrm{Var}(X)=n\\dfrac{K}{N}\\left(1-\\dfrac{K}{N}\\right)\\dfrac{N-n}{N-1}=5(0.4)(0.6)\\dfrac{15}{19}=1.2\\cdot\\dfrac{15}{19}\\approx 0.9474$.",
      "tag": "Hypergeometric"
    },
    {
      "front": "Give the pmf, mean, variance, and MGF of a Poisson$(\\lambda)$ random variable.",
      "back": "pmf: $p(x)=\\dfrac{e^{-\\lambda}\\lambda^{x}}{x!}$ for $x=0,1,2,\\dots$.\nMean: $E[X]=\\lambda$.\nVariance: $\\mathrm{Var}(X)=\\lambda$ (mean equals variance).\nMGF: $M_X(t)=\\exp[\\lambda(e^{t}-1)]$.",
      "tag": "Poisson"
    },
    {
      "front": "Claims arrive Poisson with mean $\\lambda=3$ per day. Find the probability of at least $2$ claims on a given day.",
      "back": "$\\Pr(X \\geq 2)=1-\\Pr(X=0)-\\Pr(X=1)$.\n$\\Pr(X=0)=e^{-3}\\approx 0.049787$; $\\Pr(X=1)=3e^{-3}\\approx 0.149361$.\n$\\Pr(X \\geq 2)=1-0.049787-0.149361\\approx 0.8009$.",
      "tag": "Poisson"
    },
    {
      "front": "For $X\\sim\\text{Poisson}(2.5)$, compute $\\Pr(X=3)$.",
      "back": "$\\Pr(X=3)=\\dfrac{e^{-2.5}(2.5)^{3}}{3!}=\\dfrac{e^{-2.5}\\cdot 15.625}{6}$.\n$e^{-2.5}\\approx 0.082085$, so $\\Pr(X=3)\\approx \\dfrac{0.082085\\cdot 15.625}{6}\\approx 0.2138$.",
      "tag": "Poisson"
    },
    {
      "front": "For $X\\sim\\text{Poisson}(1.5)$, find $\\Pr(X \\leq 2)$.",
      "back": "Sum the first three terms:\n$\\Pr(X \\leq 2)=e^{-1.5}\\left(1+1.5+\\dfrac{1.5^{2}}{2}\\right)=e^{-1.5}(1+1.5+1.125)=e^{-1.5}(3.625)$.\n$e^{-1.5}\\approx 0.223130$, so $\\Pr(X \\leq 2)\\approx 0.8088$.",
      "tag": "Poisson"
    },
    {
      "front": "Accidents occur Poisson at $\\lambda=4$ per week. Find the probability of exactly $6$ accidents over a two-week period.",
      "back": "Over two weeks the rate scales: $\\lambda'=4\\cdot 2=8$.\n$\\Pr(X=6)=\\dfrac{e^{-8}8^{6}}{6!}=\\dfrac{e^{-8}\\cdot 262144}{720}$.\n$e^{-8}\\approx 0.000335463$, so $\\Pr(X=6)\\approx 0.1221$.",
      "tag": "Poisson"
    },
    {
      "front": "For a Poisson$(\\lambda)$ variable, what is the factorial moment $E[X(X-1)]$, and use it to find $E[X^{2}]$ when $\\lambda=4$.",
      "back": "The Poisson factorial moment is $E[X(X-1)]=\\lambda^{2}$.\nSo $E[X^{2}]=E[X(X-1)]+E[X]=\\lambda^{2}+\\lambda$.\nWith $\\lambda=4$: $E[X^{2}]=16+4=20$ (consistent with $\\mathrm{Var}=E[X^{2}]-(E[X])^{2}=20-16=4=\\lambda$).",
      "tag": "Poisson"
    },
    {
      "front": "State the Poisson approximation to the binomial (Poisson as the limit of the binomial).",
      "back": "If $n\\to\\infty$ and $p\\to 0$ with the product $np\\to\\lambda$ held fixed, then $\\text{Binomial}(n,p)\\to\\text{Poisson}(\\lambda)$.\nIn practice it is accurate when $n$ is large and $p$ small (rare events), setting $\\lambda=np$.",
      "tag": "Poisson"
    },
    {
      "front": "A defect occurs independently in each of $200$ items with probability $0.01$. Use the Poisson approximation to estimate the probability of no defects.",
      "back": "Set $\\lambda=np=200(0.01)=2$.\n$\\Pr(X=0)\\approx e^{-\\lambda}=e^{-2}\\approx 0.1353$.\n(The exact binomial value $(0.99)^{200}\\approx 0.1340$ is very close, confirming the approximation.)",
      "tag": "Poisson"
    },
    {
      "front": "How does the MGF prove that the sum of independent Poisson variables is again Poisson?",
      "back": "For independent $X_i\\sim\\text{Poisson}(\\lambda_i)$, MGFs multiply:\n$M_{\\sum X_i}(t)=\\prod_i \\exp[\\lambda_i(e^{t}-1)]=\\exp\\!\\Big[\\big(\\sum_i\\lambda_i\\big)(e^{t}-1)\\Big]$.\nThis is the MGF of $\\text{Poisson}\\big(\\sum_i\\lambda_i\\big)$, so the sum is Poisson with the rates added.",
      "tag": "Poisson"
    },
    {
      "front": "State the thinning (splitting) property of a Poisson process.",
      "back": "If events occur Poisson$(\\lambda)$ and each is independently classified type-1 with probability $p$ (type-2 with $1-p$), then the type-1 count is $\\text{Poisson}(\\lambda p)$ and the type-2 count is $\\text{Poisson}(\\lambda(1-p))$.\nMoreover these two counts are **independent** of each other.",
      "tag": "Poisson"
    },
    {
      "front": "A call center gets calls Poisson at $\\lambda=20$/hour; each call independently leads to a sale with probability $0.25$. Find the distribution and mean of hourly sales.",
      "back": "By thinning, sales $\\sim\\text{Poisson}(\\lambda p)=\\text{Poisson}(20\\cdot 0.25)=\\text{Poisson}(5)$.\nMean sales per hour $=5$ (variance also $5$).",
      "tag": "Poisson"
    },
    {
      "front": "For a general random variable, how do the linear-transform rules $E[aX+b]$ and $\\mathrm{Var}(aX+b)$ work?",
      "back": "Expectation is linear: $E[aX+b]=aE[X]+b$.\nVariance ignores the shift and squares the scale: $\\mathrm{Var}(aX+b)=a^{2}\\mathrm{Var}(X)$, so $\\mathrm{SD}(aX+b)=|a|\\,\\sigma$.\nThese hold for any distribution, discrete or continuous.",
      "tag": "Moments/MGF"
    },
    {
      "front": "How do you recover moments from an MGF, and what is $M_X(0)$ for any valid MGF?",
      "back": "Differentiate and evaluate at $0$: $E[X^{n}]=M_X^{(n)}(0)$, the $n$-th derivative at $t=0$.\nIn particular $E[X]=M_X'(0)$ and $E[X^{2}]=M_X''(0)$.\nEvery MGF satisfies $M_X(0)=1$ (since $e^{0\\cdot X}=1$). The MGF, when it exists, uniquely determines the distribution.",
      "tag": "Moments/MGF"
    },
    {
      "front": "An MGF is $M_X(t)=\\exp[5(e^{t}-1)]$. Identify the distribution and state $E[X]$ and $\\mathrm{Var}(X)$.",
      "back": "This matches the Poisson MGF $\\exp[\\lambda(e^{t}-1)]$ with $\\lambda=5$, so $X\\sim\\text{Poisson}(5)$.\nTherefore $E[X]=\\lambda=5$ and $\\mathrm{Var}(X)=\\lambda=5$.",
      "tag": "Moments/MGF"
    },
    {
      "front": "An MGF is $M_X(t)=(0.7+0.3e^{t})^{6}$. Identify the distribution and compute its mean and variance.",
      "back": "This matches the binomial MGF $(q+pe^{t})^{n}$ with $n=6$, $p=0.3$, $q=0.7$, so $X\\sim\\text{Binomial}(6,0.3)$.\nMean: $E[X]=np=1.8$.\nVariance: $\\mathrm{Var}(X)=npq=6(0.3)(0.7)=1.26$.",
      "tag": "Moments/MGF"
    },
    {
      "front": "An MGF is $M_X(t)=\\dfrac{0.4e^{t}}{1-0.6e^{t}}$. Identify the distribution and give $E[X]$.",
      "back": "This matches the geometric MGF (convention A) $\\dfrac{pe^{t}}{1-qe^{t}}$ with $p=0.4$, $q=0.6$.\nSo $X\\sim\\text{Geometric}(0.4)$ on $\\{1,2,\\dots\\}$, giving $E[X]=\\dfrac{1}{p}=2.5$.",
      "tag": "Moments/MGF"
    }
  ]
}