Probabilistic Methods — Interactive Solver

Task 1 — PMF Table (Custom Discrete)

A discrete random variable $X$ has the following probability mass function: $P(X=-2)=0.10$ , $P(X=0)=0.25$ , $P(X=1)=0.30$ , $P(X=3)=0.20$ , $P(X=5)=0.15$ . Verify the PMF sums to 1 and derive the CDF $F(x)=P(X\leq x)$ .

PMF Verification

Sum all probabilities to confirm they form a valid PMF.

\sum_x P(X=x)=0.10+0.25+0.30+0.20+0.15=1.00\checkmark

PMF (Tabular Form)

The probability mass function assigns mass only at the five support points.

P(X=x)=\begin{cases}0.10 & x=-2\\0.25 & x=0\\0.30 & x=1\\0.20 & x=3\\0.15 & x=5\\0 & \text{otherwise}\end{cases}

CDF Definition

The cumulative distribution function $F(x)=P(X\leq x)$ is a right-continuous step function that jumps by $P(X=x)$ at each support point.

CDF (Step Function)

Accumulating the PMF values from left to right gives the CDF.

F(x)=\begin{cases}0 & x<-2\\0.10 & -2\leq x<0\\0.35 & 0\leq x<1\\0.65 & 1\leq x<3\\0.85 & 3\leq x<5\\1.00 & x\geq 5\end{cases}

Reading the CDF

For example, $P(X\leq 1)=F(1)=0.65$ and $P(0<X\leq 3)=F(3)-F(0)=0.85-0.35=0.50$ .

P(a<X\leq b)=F(b)-F(a)

Answer

F(x) is a right-continuous step function with jumps of 0.10, 0.25, 0.30, 0.20, 0.15 at x = −2, 0, 1, 3, 5 respectively.

Explanation tip: A valid PMF requires all probabilities to be non-negative and sum exactly to 1. The CDF is non-decreasing, right-continuous, with

\lim_{x\to-\infty}F(x)=0

and

\lim_{x\to+\infty}F(x)=1

.

Task 2 — CDF Table → PMF

A discrete random variable $X$ has CDF values: $F(-1)=0.15$ , $F(0)=0.35$ , $F(2)=0.60$ , $F(4)=0.85$ , $F(6)=1.00$ . Recover the PMF from these jump sizes.

Key Property

For a discrete distribution, the PMF at each support point equals the jump (upward step) of the CDF at that point.

P(X=x_i)=F(x_i)-F(x_i^-)

Recovering PMF Values

Compute each jump: since $F$ is 0 before the first support point, the jump at $x=-1$ is $F(-1)-0=0.15$ .

\begin{aligned}P(X=-1)&=F(-1)-0=0.15\\P(X=0)&=F(0)-F(-1)=0.35-0.15=0.20\\P(X=2)&=F(2)-F(0)=0.60-0.35=0.25\\P(X=4)&=F(4)-F(2)=0.85-0.60=0.25\\P(X=6)&=F(6)-F(4)=1.00-0.85=0.15\end{aligned}

PMF (Recovered)

The resulting PMF assigns probabilities only at the five support points.

P(X=x)=\begin{cases}0.15 & x=-1\\0.20 & x=0\\0.25 & x=2\\0.25 & x=4\\0.15 & x=6\\0 & \text{otherwise}\end{cases}

Verify Sum

Check that the recovered probabilities form a valid PMF.

0.15+0.20+0.25+0.25+0.15=1.00\checkmark

Right-Continuity Note

A discrete CDF is right-continuous: $F(x)$ includes the mass at $x$ itself. Between support points, $F$ is flat (constant). This means $P(X=x)=F(x)-\lim_{t\uparrow x}F(t)$ .

Answer

PMF: P(X=−1)=0.15, P(X=0)=0.20, P(X=2)=0.25, P(X=4)=0.25, P(X=6)=0.15.

Explanation tip: The jump size at each point of increase of the CDF directly gives the probability mass there. Flat segments of the CDF correspond to values with zero probability.

Task 3 — Binomial Distribution

For $X\sim\text{Bin}(n,p)$ , state the PMF and CDF, compute a worked example with $n=10$ , $p=0.3$ , and explore how the shape changes with parameters.

n = 10p = 0.3

PMF

CDF

PMF

The probability of exactly $k$ successes in $n$ independent Bernoulli trials, each with success probability $p$ .

P(X=k)=\binom{n}{k}p^k(1-p)^{n-k},\quad k=0,1,\ldots,n

CDF

The cumulative probability up to and including $k$ successes.

F(k)=P(X\leq k)=\sum_{i=0}^{k}\binom{n}{i}p^i(1-p)^{n-i}

Mean and Variance

Key moments of the Binomial distribution.

\mu=np,\quad \sigma^2=np(1-p)

Worked Example

With $n=10$ , $p=0.3$ : compute $P(X=3)$ and $P(X\leq 3)$ .

P(X=3)=\binom{10}{3}(0.3)^3(0.7)^7=120\cdot0.027\cdot0.0824\approx0.2668

CDF at k=3

Summing from $k=0$ to $k=3$ .

F(3)=\sum_{i=0}^{3}\binom{10}{i}(0.3)^i(0.7)^{10-i}\approx0.6496

Applications

Quality control (defective items in a batch), clinical trials (patients responding to treatment), poll surveys (voters supporting a candidate).

Answer

P(X=3) ≈ 0.2668; F(3) ≈ 0.6496 for Bin(10, 0.3).

Explanation tip: Use the slider to compare Bin(10,0.3), Bin(10,0.5), Bin(10,0.7) — the mean sits near

np

. As

p\to0.5

the distribution becomes symmetric.

Task 4 — Geometric Distribution

For $X\sim\text{Geo}(p)$ , where $X$ counts the trial on which the first success occurs, state the PMF and CDF, and compute a worked example with $p=0.2$ .

p = 0.2

PMF

CDF

PMF

The probability that the first success occurs on trial $k$ (after $k-1$ failures).

P(X=k)=(1-p)^{k-1}p,\quad k=1,2,3,\ldots

CDF

The probability that the first success occurs by trial $k$ (closed-form, no infinite sum needed).

F(k)=P(X\leq k)=1-(1-p)^k,\quad k=1,2,3,\ldots

Mean and Variance

Key moments of the Geometric distribution.

\mu=\frac{1}{p},\quad \sigma^2=\frac{1-p}{p^2}

Worked Example

With $p=0.2$ : probability the first success occurs on trial 4, and by trial 4.

P(X=4)=(0.8)^3(0.2)=0.512\cdot0.2=0.1024

CDF at k=4

Probability of getting the first success within 4 trials.

F(4)=1-(0.8)^4=1-0.4096=0.5904

Memoryless Property

The Geometric distribution is memoryless: past failures do not affect future probabilities.

P(X>m+n\mid X>m)=P(X>n)

Applications

Number of attempts until the first success: number of calls until a sale, number of rolls until a six, or number of tests until a defect is found.

Answer

P(X=4) = 0.1024; F(4) = 0.5904 for Geo(0.2). Expected first success at trial 5.

Explanation tip: As

p

increases, the distribution concentrates near

k=1

. The Geometric distribution is the only discrete memoryless distribution.

Task 5 — Poisson Distribution

For $X\sim\text{Pois}(\lambda)$ , state the PMF and CDF, compute a worked example with $\lambda=3$ , and identify its key applications.

λ = 3

PMF

CDF

PMF

The probability of observing exactly $k$ events in a fixed interval when events occur independently at a constant average rate $\lambda$ .

P(X=k)=\frac{e^{-\lambda}\lambda^k}{k!},\quad k=0,1,2,\ldots

CDF

Cumulative probability of at most $k$ events (regularised incomplete gamma function).

F(k)=P(X\leq k)=e^{-\lambda}\sum_{i=0}^{k}\frac{\lambda^i}{i!}=\frac{\Gamma(k+1,\lambda)}{k!}

Mean and Variance

Notably, both the mean and variance equal $\lambda$ .

\mu=\lambda,\quad \sigma^2=\lambda

Worked Example

With $\lambda=3$ : find $P(X=2)$ and $P(X\leq 2)$ .

P(X=2)=\frac{e^{-3}\cdot3^2}{2!}=\frac{0.04979\cdot9}{2}\approx0.2240

CDF at k=2

Sum of $P(X=0)$ , $P(X=1)$ , and $P(X=2)$ .

F(2)=e^{-3}\left(1+3+\frac{9}{2}\right)=e^{-3}\cdot8.5\approx0.4232

Applications

Customer arrivals per hour, number of typos per page, radioactive decay counts per second, number of insurance claims per day.

Answer

P(X=2) ≈ 0.2240; F(2) ≈ 0.4232 for Pois(3).

Explanation tip: The Poisson distribution arises as the limit of Bin

(n,p)

when

n\to\infty

,

p\to0

, and

np\to\lambda

. Larger

\lambda

shifts and spreads the distribution rightward.

Task 6 — Hypergeometric Distribution

For $X\sim\text{Hyper}(N,K,n)$ , where $N$ is the population size, $K$ the number of success states, and $n$ the number of draws (without replacement), state the PMF and compute an example with $N=20$ , $K=5$ , $n=4$ .

N = 20K = 5n = 4

PMF

CDF

PMF

The probability of drawing exactly $k$ successes in $n$ draws without replacement from a population of $N$ items containing $K$ successes.

P(X=k)=\frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}},\quad \max(0,n+K-N)\leq k\leq\min(n,K)

CDF

Cumulative sum of the PMF over valid support values.

F(k)=\sum_{i=0}^{k}\frac{\binom{K}{i}\binom{N-K}{n-i}}{\binom{N}{n}}

Mean and Variance

The mean mirrors the Binomial with $p=K/N$ ; the variance includes a finite-population correction factor.

\mu=n\frac{K}{N},\quad \sigma^2=n\frac{K}{N}\cdot\frac{N-K}{N}\cdot\frac{N-n}{N-1}

Worked Example

With $N=20$ , $K=5$ , $n=4$ : probability of drawing exactly 2 successes.

P(X=2)=\frac{\binom{5}{2}\binom{15}{2}}{\binom{20}{4}}=\frac{10\cdot105}{4845}=\frac{1050}{4845}\approx0.2167

CDF at k=2

Sum probabilities for $k=0,1,2$ .

P(X=0)=\frac{\binom{5}{0}\binom{15}{4}}{\binom{20}{4}}=\frac{1365}{4845}\approx0.2817

Relation to Binomial

When the population $N$ is very large relative to $n$ , sampling without replacement is nearly the same as sampling with replacement.

\text{Hyper}(N,K,n)\xrightarrow{N\to\infty}\text{Bin}\!\left(n,\frac{K}{N}\right)

Applications

Quality sampling (defective items in a batch), ecology (capture-recapture), card games (drawing from a deck), lottery odds.

Answer

P(X=2) ≈ 0.2167 for Hyper(20, 5, 4). Mean = 4×(5/20) = 1.

Explanation tip: Unlike the Binomial, trials in the Hypergeometric are dependent (no replacement). The finite-population correction

\frac{N-n}{N-1}

always reduces variance below the Binomial variance.

Task 7 — Negative Binomial Distribution

For $X\sim\text{NegBin}(r,p)$ , where $X$ is the trial number on which the $r$ -th success occurs, state the PMF and explore its connection to the Geometric distribution, using $r=3$ , $p=0.4$ .

r = 3p = 0.4

PMF

CDF

PMF

The probability that the $r$ -th success occurs on trial $k$ (exactly $r-1$ successes in the first $k-1$ trials, then a success on trial $k$ ).

P(X=k)=\binom{k-1}{r-1}p^r(1-p)^{k-r},\quad k=r,r+1,r+2,\ldots

CDF

Cumulative probability that the $r$ -th success occurs by trial $k$ , expressible via the regularised incomplete beta function.

F(k)=I_p(r,\,k-r+1)=\sum_{i=r}^{k}\binom{i-1}{r-1}p^r(1-p)^{i-r}

Mean and Variance

Key moments of the Negative Binomial distribution.

\mu=\frac{r}{p},\quad \sigma^2=\frac{r(1-p)}{p^2}

Connection to Geometric

When $r=1$ , the Negative Binomial reduces to the Geometric distribution.

P(X=k\mid r=1)=(1-p)^{k-1}p=\text{Geo}(p)

Worked Example

With $r=3$ , $p=0.4$ : probability the 3rd success occurs on trial 6.

P(X=6)=\binom{5}{2}(0.4)^3(0.6)^3=10\cdot0.064\cdot0.216=0.1382

Applications

Number of sales calls to close 3 deals, number of experiments to achieve $r$ successes, modeling over-dispersed count data (variance $>$ mean).

Answer

P(X=6) ≈ 0.1382 for NegBin(3, 0.4). Expected trial for 3rd success = 3/0.4 = 7.5.

Explanation tip: The sum of

r

independent Geo

(p)

random variables follows NegBin

(r,p)

. Adjust the slider to see how increasing

r

shifts the mass rightward and reduces skewness.

Task 8 — Beta Distribution

For $X\sim\text{Beta}(\alpha,\beta)$ on $[0,1]$ , state the PDF and CDF, compute a numerical note, and explore how the shape parameters change the distribution.

α = 2β = 2

PDF

CDF

PDF

The Beta PDF is defined on the unit interval $[0,1]$ and controlled by two positive shape parameters $\alpha$ and $\beta$ .

f(x)=\frac{x^{\alpha-1}(1-x)^{\beta-1}}{B(\alpha,\beta)},\quad x\in[0,1]

Beta Function Normalising Constant

The Beta function $B(\alpha,\beta)$ ensures the PDF integrates to 1.

B(\alpha,\beta)=\int_0^1 t^{\alpha-1}(1-t)^{\beta-1}\,dt=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}

CDF

The CDF is the regularised incomplete Beta function $I_x(\alpha,\beta)$ .

F(x)=I_x(\alpha,\beta)=\frac{B(x;\,\alpha,\beta)}{B(\alpha,\beta)}=\frac{\int_0^x t^{\alpha-1}(1-t)^{\beta-1}\,dt}{B(\alpha,\beta)}

Mean and Variance

Key moments of the Beta distribution.

\mu=\frac{\alpha}{\alpha+\beta},\quad \sigma^2=\frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}

Worked Example

With $\alpha=2$ , $\beta=2$ : the PDF is $f(x)=6x(1-x)$ , symmetric about $x=0.5$ .

B(2,2)=\frac{\Gamma(2)\Gamma(2)}{\Gamma(4)}=\frac{1\cdot1}{6}=\frac{1}{6},\quad f(x)=6x(1-x)

Shape Varieties

Different parameter choices yield very different shapes.

\begin{aligned}&\text{Beta}(1,1): \text{Uniform}[0,1]\\&\text{Beta}(2,2): \text{symmetric, unimodal}\\&\text{Beta}(5,2): \text{right-skewed, peak toward 1}\\&\text{Beta}(0.5,0.5): \text{U-shaped, bimodal at endpoints}\end{aligned}

Applications

Modelling proportions and probabilities (e.g., click-through rates, Bayesian prior for a Binomial probability), order statistics of the uniform distribution.

Answer

For Beta(2,2): f(x) = 6x(1−x), mean = 0.5, variance = 1/20 = 0.05.

Explanation tip: The Beta distribution is the conjugate prior for the Binomial likelihood in Bayesian inference. Increasing both parameters while keeping their ratio fixed sharpens the distribution around the mean.

Task 9 — Gamma Distribution

For $X\sim\text{Gamma}(k,\theta)$ with shape $k>0$ and scale $\theta>0$ , state the PDF and CDF, give a worked example with $k=2$ , $\theta=2$ , and identify the Chi-square special case.

k = 2θ = 2

PDF

CDF

PDF

The Gamma PDF models the waiting time until the $k$ -th event in a Poisson process with rate $1/\theta$ .

f(x)=\frac{x^{k-1}e^{-x/\theta}}{\theta^k\,\Gamma(k)},\quad x\geq0

Gamma Function

The Gamma function generalises the factorial: $\Gamma(n)=(n-1)!$ for positive integers.

\Gamma(k)=\int_0^\infty t^{k-1}e^{-t}\,dt,\quad \Gamma(k+1)=k\,\Gamma(k)

CDF

The CDF is expressed through the regularised lower incomplete Gamma function $P(k,x/\theta)$ .

F(x)=P\!\left(k,\,\frac{x}{\theta}\right)=\frac{\gamma(k,\,x/\theta)}{\Gamma(k)}

Mean and Variance

Key moments of the Gamma distribution.

\mu=k\theta,\quad \sigma^2=k\theta^2

Worked Example

With $k=2$ , $\theta=2$ : the PDF simplifies to $f(x)=xe^{-x/2}/4$ , $x\geq0$ .

f(x)=\frac{x^{2-1}e^{-x/2}}{2^2\,\Gamma(2)}=\frac{xe^{-x/2}}{4\cdot1}=\frac{x\,e^{-x/2}}{4}

Mean and SD for example

Numerical values for the $k=2$ , $\theta=2$ case.

\mu=2\cdot2=4,\quad \sigma=\sqrt{2\cdot4}=\sqrt{8}\approx2.83

Chi-Square Special Case

The Chi-square distribution with $\nu$ degrees of freedom is a Gamma with $k=\nu/2$ and $\theta=2$ .

\chi^2_\nu=\text{Gamma}\!\left(\frac{\nu}{2},\,2\right)

Applications

Waiting time until $k$ events in a Poisson process, insurance claim sizes, rainfall amounts, time to failure in reliability analysis.

Answer

For Gamma(2,2): f(x) = xe^{−x/2}/4, mean = 4, SD ≈ 2.83.

Explanation tip: Exponential

(\theta)

= Gamma

(1,\theta)

. As

k

increases, the Gamma distribution becomes more symmetric and approaches a Normal shape by the Central Limit Theorem.

Task 10 — Normal Distribution

For $X\sim N(\mu,\sigma^2)$ , state the PDF and CDF, explain the zero-probability property of continuous distributions, and explore applications with the standard normal $N(0,1)$ .

μ = 0σ = 1

PDF

CDF

PDF

The Normal (Gaussian) distribution is the most important continuous distribution, characterised by its bell-shaped, symmetric PDF.

f(x)=\frac{1}{\sigma\sqrt{2\pi}}\exp\!\left(-\frac{(x-\mu)^2}{2\sigma^2}\right),\quad x\in(-\infty,+\infty)

CDF

The CDF has no closed form; it is expressed through the error function $\operatorname{erf}$ or the standard normal CDF $\Phi$ .

F(x)=\Phi\!\left(\frac{x-\mu}{\sigma}\right)=\frac{1}{2}\left[1+\operatorname{erf}\!\left(\frac{x-\mu}{\sigma\sqrt{2}}\right)\right]

Standardisation

Any Normal variable can be standardised to $Z\sim N(0,1)$ .

Z=\frac{X-\mu}{\sigma}\sim N(0,1)

Zero Probability at a Point

For ANY continuous distribution, the probability of any single exact value is zero — a single point has zero width and thus zero area under the PDF.

P(X=a)=\int_a^a f(x)\,dx=0\quad\text{for all }a\in\mathbb{R}

Interval Probabilities

Meaningful probabilities are only defined for intervals.

P(a\leq X\leq b)=\int_a^b f(x)\,dx=F(b)-F(a)

Empirical Rule

The 68-95-99.7 rule describes probability within 1, 2, 3 standard deviations.

\begin{aligned}P(\mu-\sigma\leq X\leq\mu+\sigma)&\approx0.6827\\P(\mu-2\sigma\leq X\leq\mu+2\sigma)&\approx0.9545\\P(\mu-3\sigma\leq X\leq\mu+3\sigma)&\approx0.9973\end{aligned}

Worked Example

For $X\sim N(0,1)$ : find $P(-1\leq X\leq 1)$ .

P(-1\leq X\leq 1)=\Phi(1)-\Phi(-1)=0.8413-0.1587=0.6827

Applications

Measurement error in physical experiments, human heights and weights, standardised test scores (IQ, SAT), financial returns (as a first approximation), noise in signal processing.

Answer

For N(0,1): P(−1 ≤ X ≤ 1) ≈ 0.6827. For any continuous distribution, P(X = a) = 0.

Explanation tip: The Normal distribution is justified theoretically by the Central Limit Theorem: the sum (or mean) of a large number of independent, identically distributed random variables converges to Normal regardless of the original distribution.

List 5 — PMF, PDF, and CDF

Task 1 — PMF Table (Custom Discrete)

Task 2 — CDF Table → PMF

Task 3 — Binomial Distribution

Task 4 — Geometric Distribution

Task 5 — Poisson Distribution

Task 6 — Hypergeometric Distribution

Task 7 — Negative Binomial Distribution

Task 8 — Beta Distribution

Task 9 — Gamma Distribution

Task 10 — Normal Distribution