Random Variables and Discrete Probability Distributions

Introduction to Random Variables

A random variable (RV) assigns a numerical value to each outcome of a random experiment. Think of it as a rule that converts the abstract outcome (e.g., "heads") into a number you can work with (e.g., $0$).

Discrete vs. Continuous Random Variables

Note: In the same setting (e.g., a bank), the number of customers arriving is a discrete RV, while the time between arrivals is a continuous RV.

Probability Mass Function (PMF)

For a discrete RV $X$, the probability mass function (PMF), denoted $f(x)$, gives the probability that $X$ equals exactly a particular value $x$:

$f(x) = P(X = x)$

• $f(x) \ge 0$ for all $x$
• $\sum_{\text{all }x} f(x) = 1$ (total probability mass = 1)

Why “mass”? Think of 1 unit of probability being distributed across the possible values — the PMF decides how much mass lands on each value.

Example: Fair die

$f(x) = \frac{1}{6}, \quad x = 1,2,3,4,5,6$

Plot: six bars of equal height ($1/6$).

Example: Biased die

Plot: bars of different heights (tallest at $x=1$).

Cumulative Distribution Function (CDF) and Tail Probability

The cumulative distribution function (CDF), $F(\phi)$, gives the probability that $X$ is less than or equal to $\phi$:

$F(\phi) = P(X \le \phi) = \sum_{x \le \phi} f(x)$

The tail probability (complement) is:

$\bar{F}(x) = P(X > x) = 1 - F(x)$

Worked example: Fair die

• $F(5) = P(X \le 5) = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{5}{6}$
• $P(X > 3) = 1 - F(3) = 1 - \frac{3}{6} = \frac{1}{2}$
• $P(\text{even}) = P(X=2,4,6) = \frac{3}{6} = \frac{1}{2}$
• $P(2 \le X \le 5) = P(X=2,3,4,5) = \frac{4}{6} = \frac{2}{3}$

Worked example: Biased die (probabilities as above)

• $P(X=2) = f(2) = \frac{1}{6}$
• $P(X \le 5) = f(1)+f(2)+f(3)+f(4)+f(5) = \frac{1}{3}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{12} = \frac{11}{12}$ (Equivalently, $1 - f(6) = 1 - \frac{1}{12} = \frac{11}{12}$)

Expected Value, Variance, and Standard Deviation

Expected value $E[X]$ is the probability‑weighted average of all possible values:

$E[X] = \sum_{\text{all }x} x \cdot f(x)$

• Provides the center of the distribution.
• It need not be a value the RV can actually take (e.g., fair die: $E[X] = 3.5$, yet you never roll a 3.5).

Variance $\text{Var}(X)$ measures the spread — the probability‑weighted average of squared deviations from the mean:

$\text{Var}(X) = \sum_{\text{all }x} \big(x - E[X]\big)^2 \cdot f(x)$

Standard deviation $\sigma_X$ is the square root of the variance (same units as X):

$\sigma_X = \sqrt{\text{Var}(X)}$

Worked example: Fair die

$E[X] = 1\cdot\frac{1}{6} + 2\cdot\frac{1}{6} + \cdots + 6\cdot\frac{1}{6} = \frac{21}{6} = 3.5$

$$\begin{aligned} \text{Var}(X) &= (1-3.5)^2\cdot\frac{1}{6} + (2-3.5)^2\cdot\frac{1}{6} + \cdots + (6-3.5)^2\cdot\frac{1}{6} \\[2pt] &= \frac{35}{12} \approx 2.92 \end{aligned}$$

$\sigma_X = \sqrt{\frac{35}{12}} \approx 1.71$

Worked example: Biased die

From the PMF table above:

$$\begin{aligned} E[X] &= 1\cdot\frac{1}{3} + 2\cdot\frac{1}{6} + 3\cdot\frac{1}{6} + 4\cdot\frac{1}{6} + 5\cdot\frac{1}{12} + 6\cdot\frac{1}{12} \\ &= \frac{33}{12} = 2.75 \end{aligned}$$ $$\begin{aligned} \text{Var}(X) &= (1-2.75)^2\cdot\frac{1}{3} + (2-2.75)^2\cdot\frac{1}{6} + (3-2.75)^2\cdot\frac{1}{6} \\ &\quad + (4-2.75)^2\cdot\frac{1}{6} + (5-2.75)^2\cdot\frac{1}{12} + (6-2.75)^2\cdot\frac{1}{12} \\ &= 2.6875 \approx 2.69 \end{aligned}$$

Exam tip: For any discrete RV, always check that the probabilities sum to 1 before computing expected value or variance. Also remember: $E[X]$ may be a number the RV never actually takes — it describes the long‑run average, not a possible outcome.

Key takeaways

• A random variable maps random outcomes to numbers; discrete RVs take countable values, continuous RVs take any value in an interval.
• The PMF $f(x)$ gives $P(X=x)$ and sums to 1.
• The CDF $F(x)$ gives $P(X\le x)$; tail probability $P(X>x) = 1-F(x)$.
• Expected value $E[X] = \sum x \cdot f(x)$ is the probability‑weighted mean.
• Variance $\text{Var}(X) = \sum (x-E[X])^2 f(x)$; standard deviation = $\sqrt{\text{Var}(X)}$.
• Both fair and biased dice illustrate that probabilities change the center and spread, even though the possible values stay the same.

Key Concepts: PMF, CDF, Tail Probability, Expectation, Variance

For a discrete random variable $X$:

• Probability Mass Function (PMF): $f(x) = P(X = x)$. Non‑negative and sums to 1 over all $x$.
• Cumulative Distribution Function (CDF): $F(x) = P(X \leq x) = \sum_{t \leq x} f(t)$.
• Tail probability: $\bar{F}(x) = P(X > x) = 1 - F(x)$.
• Expected value: $\mu = E[X] = \sum x \cdot f(x)$.
• Variance: $\sigma^2 = \text{Var}(X) = \sum (x - \mu)^2 f(x) = E[X^2] - \mu^2$; standard deviation $\sigma = \sqrt{\sigma^2}$.

From Data to PMF

When the true PMF is unknown, estimate it using relative frequencies from observed data:

$$f(x) \approx \frac{\text{number of occurrences of } x}{\text{total observations}}.$$

This gives a valid PMF because frequencies are non‑negative and sum to 1.

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Example 1: Biased Die (Reviewing PMF, CDF, Tail)

A biased die has PMF:

$$f(1)=f(2)=f(3)=f(4)=\frac{1}{6},\quad f(5)=f(6)=\frac{1}{12}.$$

• Probability of even value: $P(\text{even}) = f(2)+f(4)+f(6) = \frac{1}{6}+\frac{1}{6}+\frac{1}{12} = \frac{5}{12}$.
• Probability of values between 2 and 5 inclusive: $P(2 \leq X \leq 5) = f(2)+f(3)+f(4)+f(5) = \frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{12} = \frac{7}{12}$.
• Tail probability $P(X > 3)$: $P(X > 3) = f(4)+f(5)+f(6) = \frac{1}{6}+\frac{1}{12}+\frac{1}{12} = \frac{1}{3}$, or $1 - F(3) = 1 - (f(1)+f(2)+f(3)) = 1 - \frac{1}{2} = \frac{1}{3}$.

Approach to any question:

1. Translate the question into an event.
2. Identify the range of $x$ values that satisfy the event.
3. Sum the corresponding $f(x)$.

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Example 2: Defective Pumps at Kirloskar Factory

Setup: $X$ = number of defective pumps per day. Data from 200 days:

Questions answered using the PMF:

• $P(\text{exactly 2}) = f(2) = 0.20$.
• $P(\leq 3) = f(0)+f(1)+f(2)+f(3) = 0.40+0.25+0.20+0.05 = 0.90$.
• Expected value: $$E[X] = \sum x f(x) = 0(0.40) + 1(0.25) + 2(0.20) + 3(0.05) + 4(0.10) = 1.2 \text{ defective pumps/day}.$$
• Variance and standard deviation: $$\begin{aligned} \text{Var}(X) &= (0-1.2)^2(0.40) + (1-1.2)^2(0.25) + (2-1.2)^2(0.20) + (3-1.2)^2(0.05) + (4-1.2)^2(0.10) \\ &= 1.66, \\ \sigma &= \sqrt{1.66} \approx 1.29. \end{aligned}$$

Exam tip: The steps are identical for any discrete distribution: obtain $f(x)$ (here from data), then compute probabilities via summation, expectation as a weighted sum, and variance as the weighted squared deviation.

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Example 3: Late Take-offs at Nagpur Airport

Setup: $X$ = number of planes taking off late between 9 pm and 11 pm. Data from 30 days:

The PMF is derived from the given statements: $f(0)=0.10$, $f(1)=0.20$, $P(\geq 2)=0.70$, $P(\leq 3)=0.80$, $E[X]=2.20$, $\text{Var}(X)=1.56$.

Questions answered:

• $P(\text{2 or more}) = 1 - f(0) - f(1) = 1 - 0.10 - 0.20 = 0.70$.
• $P(\text{4 or less}) = F(3) = f(0)+f(1)+f(2)+f(3) = 0.10+0.20+0.30+0.20 = 0.80$.
• Expected value: $$E[X] = 0(0.10) + 1(0.20) + 2(0.30) + 3(0.20) + 4(0.20) = 2.20 \text{ late planes}.$$
• Variance and standard deviation: $$\begin{aligned} E[X^2] &= 0^2(0.10) + 1^2(0.20) + 2^2(0.30) + 3^2(0.20) + 4^2(0.20) = 6.40, \\ \text{Var}(X) &= 6.40 - (2.20)^2 = 1.56, \\ \sigma &= \sqrt{1.56} \approx 1.25. \end{aligned}$$

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Key Takeaways

• PMF from data: relative frequencies estimate probabilities; sum to 1.
• Calculating probabilities: add $f(x)$ for the event’s $x$ range; use $1-F(x)$ for tail probabilities.
• Expectation: weighted average $\sum x f(x)$.
• Variance & std dev: weighted average of squared deviations; $\sigma$ gives spread in original units.
• Procedure is universal – same steps for any discrete random variable once $f(x)$ is known.

Linear Combinations of Random Variables

A linear combination of random variables is a new random variable formed by multiplying each original variable by a constant and summing them: $Y = a_1 X_1 + a_2 X_2 + \dots + a_n X_n$. Why does this matter? Because quantities like total cost, total profit, portfolio returns, and sample averages are all linear combinations. Understanding their expectation and variance lets us predict average outcomes and measure risk.

Linear Transformation of a Single Variable

If $Y = aX + b$ (a linear function of a single random variable $X$), then

• Expectation: $E(Y) = a\,E(X) + b$ Intuition: scaling and shifting the variable scales and shifts its mean.
• Variance: $\operatorname{Var}(Y) = a^2 \operatorname{Var}(X)$ Intuition: shifting ($+b$) does not affect spread; scaling ($a$) multiplies spread by $a^2$.

Example 1: Temperature conversion Ideal frying temperature in Celsius: $X$ with $E(X)=160$, $\operatorname{sd}(X)=5$. Convert to Fahrenheit: $Y = \frac{9}{5}X + 32$. $E(Y)=\frac{9}{5}\cdot160+32=320,\quad \operatorname{Var}(Y)=\left(\frac{9}{5}\right)^2\cdot25=81,\quad \operatorname{sd}(Y)=9.$

Example 2: Real estate price (rupees to dollars) Price in rupees: $X$ with $E(X)=36,\!000$, $\operatorname{sd}(X)=9,\!600$. Exchange rate $1\,\text{USD}=80\,\text{INR}$ → $Y = \frac{1}{80}X + 0$. $E(Y)=\frac{36,\!000}{80}=450,\quad \operatorname{Var}(Y)=\frac{1}{80^2}\cdot9,\!600^2=14,\!400,\quad \operatorname{sd}(Y)=120.$

Sum of Two Random Variables

For $Y = X_1 + X_2$:

• Expectation: $E(Y) = E(X_1) + E(X_2)$ (always additive).
• Variance: $\operatorname{Var}(Y) = \operatorname{Var}(X_1) + \operatorname{Var}(X_2) + 2\,\operatorname{Cov}(X_1, X_2)$ where $\operatorname{Cov}(X_1, X_2) = \rho_{12}\,\sigma_1\sigma_2$.

If $X_1$ and $X_2$ are independent, then $\rho_{12}=0$ and: $\operatorname{Var}(Y) = \operatorname{Var}(X_1) + \operatorname{Var}(X_2).$

Example: Store sales Store A: $E(X)=7,\!000$, $\operatorname{sd}(X)=1,\!500$ Store B: $E(Y)=8,\!000$, $\operatorname{sd}(Y)=2,\!000$ Independent.

• Total sales $S = X+Y$: $E(S)=15,\!000$, $\operatorname{Var}(S)=1,\!500^2+2,\!000^2=6,\!250,\!000$, $\operatorname{sd}(S)=2,\!500$.
• Difference $D = X-Y$: $E(D)=-1,\!000$, $\operatorname{Var}(D)=1,\!500^2+2,\!000^2=6,\!250,\!000$, $\operatorname{sd}(D)=2,\!500$ (same variance because $(-1)^2=1$).

Key insight: For independent variables, the variance of a difference is the sum of variances, not the difference.

General Linear Combination

Let $Y = a_1 X_1 + a_2 X_2 + \dots + a_n X_n$. Then:

• Expectation (always): $E(Y) = a_1 E(X_1) + a_2 E(X_2) + \dots + a_n E(X_n).$

• Variance (if independent): $\operatorname{Var}(Y) = a_1^2 \operatorname{Var}(X_1) + a_2^2 \operatorname{Var}(X_2) + \dots + a_n^2 \operatorname{Var}(X_n).$

• Variance (if not independent): add pairwise covariance terms: $\operatorname{Var}(Y) = \sum_{i=1}^n a_i^2 \operatorname{Var}(X_i) + 2\sum_{i<j} a_i a_j \operatorname{Cov}(X_i, X_j).$

Sum and Difference of Two Independent Identical Variables

If $X_1$ and $X_2$ are independent, $E(X_1)=E(X_2)=\mu$, $\operatorname{Var}(X_1)=\operatorname{Var}(X_2)=\sigma^2$, then:

Both have the same variance — the sign of the constant does not affect the squared scaling.

Portfolio of Two Stocks (Correlated)

Shweta Kulkarni invests in Siemens ($X$) and Trent ($Y$). $E(X)=3\%$, $\operatorname{sd}(X)=1.5\%$; $E(Y)=5\%$, $\operatorname{sd}(Y)=3\%$; correlation $\rho=-0.6$.

Portfolio return: $P = wX + (1-w)Y$, where $w$ = fraction in Siemens.

Formulas: $E(P) = w\cdot3 + (1-w)\cdot5$ $\operatorname{Var}(P) = w^2(1.5^2) + (1-w)^2(3^2) + 2w(1-w)(-0.6)(1.5)(3)$

For $w=0.5$ (equal split): $E(P)=4\%,\quad \operatorname{Var}(P)=1.4625,\quad \operatorname{sd}(P)\approx1.21\%.$

Exploring other splits:

The 90-10 portfolio is dominated by the 50-50 portfolio: same risk but lower return. After removing dominated choices, the remaining set of portfolios are Pareto optimal — each offers a different risk-return trade-off. Shweta’s optimal split depends on her risk appetite.

Exam tip: Negative correlation reduces portfolio variance. Even without correlation, diversification often lowers risk relative to a single asset.

Sum of $n$ Independent Identical Variables

Let $X_1, X_2, \dots, X_n$ be i.i.d. with mean $\mu$ and variance $\sigma^2$. Define $S = \sum_{i=1}^n X_i$. Then:

$E(S) = n\mu,\qquad \operatorname{Var}(S) = n\sigma^2.$

Example: Luggage weight Per passenger: $\mu=30$ kg, $\sigma=3$ kg, $n=81$ independent. $E(\text{total})=81\cdot30=2,\!430$ kg, $\operatorname{sd}(\text{total})=\sqrt{81\cdot9}=27$ kg.

Average of $n$ Independent Identical Variables

Define $\bar{X} = \frac{1}{n}\sum_{i=1}^n X_i$ (the sample mean). Then:

$E(\bar{X}) = \mu,\qquad \operatorname{Var}(\bar{X}) = \frac{\sigma^2}{n}, \qquad \operatorname{sd}(\bar{X}) = \frac{\sigma}{\sqrt{n}}.$

Example: Exam scores Per student: $\mu=70$, $\sigma=20$, $n=25$ independent. $E(\bar{X})=70$, $\operatorname{Var}(\bar{X})=\frac{400}{25}=16$, $\operatorname{sd}(\bar{X})=4$.

Key insight: Averaging reduces variance — the more data points, the more precise the estimate of the mean.

Coefficient of Variation

The coefficient of variation (CV) is a standardized measure of dispersion: $\text{CV} = \frac{\text{standard deviation}}{\text{mean}}.$

It expresses risk relative to the average — useful for comparing variability across different scales.

• Luggage total: $\text{CV} = \frac{27}{2430} \approx 0.011 = 1\%$.
• Exam average: $\text{CV} = \frac{4}{70} \approx 0.057 = 5.7\%$.

Key takeaways

• $E(aX+b) = aE(X)+b$; $\operatorname{Var}(aX+b) = a^2\operatorname{Var}(X)$.
• For sums: expectations add always; variances add only if independent; with dependence, include covariance.
• Portfolio return variance is reduced by negative correlation between assets.
• The sample mean $\bar{X}$ has variance $\sigma^2/n$ — a foundation for statistical inference.
• Coefficient of variation standardizes dispersion for comparing risk across different means.

Recap: Discrete Random Variables

A discrete random variable $X$ takes only integer values (e.g., 0, 1, 2, …). Its behaviour is fully described by the probability mass function (PMF) $f(x) = P(X = x)$. From the PMF we derive:

• Cumulative distribution function (CDF): $F(x) = P(X \leq x) = \sum_{t \leq x} f(t)$
• Expectation (mean): $E[X] = \sum_{x} x \, f(x)$
• Variance: $\text{Var}(X) = \sum_{x} (x - E[X])^2 \, f(x)$
• Standard deviation: $\text{SD}(X) = \sqrt{\text{Var}(X)}$

Linear Combinations of Random Variables

For constants $a_1, a_2, \ldots, a_n$ and random variables $X_1, X_2, \ldots, X_n$, define $Y = a_1 X_1 + a_2 X_2 + \cdots + a_n X_n$. Then:

• Expectation: $E[Y] = a_1 E[X_1] + a_2 E[X_2] + \cdots + a_n E[X_n]$
• Variance: $\text{Var}(Y) = a_1^2 \text{Var}(X_1) + a_2^2 \text{Var}(X_2) + \cdots + a_n^2 \text{Var}(X_n) + \text{extra terms}$

The “extra terms” are all pairwise covariances: $2 a_i a_j \text{Cov}(X_i, X_j)$. If the $X_i$ are independent, all covariances are zero and the variance simplifies to the sum of the squared-coefficient times variances.

Key insight: Linearity of expectation always holds; variance only simplifies under independence.

Key takeaways

• PMF gives probabilities for each outcome; CDF accumulates them.
• $E[X]$ is the probability‑weighted average; $\text{Var}(X)$ measures spread around that average.
• For linear combinations: expectation is linear, variance adds covariances (which vanish under independence).

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Bernoulli Random Variable

A Bernoulli random variable $Y$ models an experiment with exactly two outcomes: success ($Y=1$) or failure ($Y=0$). Let the success probability be $p$, and $q = 1-p$ the failure probability. The PMF is:

$$f(y) = \begin{cases} p, & y = 1\\ 1-p = q, & y = 0 \end{cases}$$

Why it matters

Before studying the Binomial distribution (which counts successes in multiple trials), we must master the Bernoulli – it is the building block. Many real‑world events can be classified as success/failure:

Expectation and Variance

Using the definitions for a discrete random variable:

$$E[Y] = (1 \times p) + (0 \times q) = p$$ $$\text{Var}(Y) = (1-p)^2 \cdot p + (0-p)^2 \cdot q = p q (p + q) = p q$$ $$\text{SD}(Y) = \sqrt{p q}$$

Exam tip: The expectation of a Bernoulli is simply $p$ – the proportion of successes. Its variance is $p(1-p)$; memorise these two formulas.

Worked Example: Biased Die

For this biased die, the event “outcome is 5 or 6” is given probability $p=\frac{1}{6}$. The individual face probabilities are not needed for the Bernoulli calculation.

• Success: die shows 5 or 6 → $p = \frac{1}{6}$
• Failure: die shows 1-4 → $q = \frac{5}{6}$

$$E[Y] = \frac{1}{6}, \quad \text{Var}(Y) = \frac{1}{6}\cdot\frac{5}{6} = \frac{5}{36}, \quad \text{SD}(Y) = \frac{\sqrt{5}}{6}$$

Examples from Earlier Contexts

In each case the Bernoulli is defined by partitioning outcomes of a known discrete distribution:

How to Create a Bernoulli from Any Discrete Random Variable

Any discrete random variable can be reduced to a Bernoulli by grouping outcomes into two categories. The examples above show exactly this: for the number of defective pumps, “≤1” becomes success and “≥2” failure, preserving the original probabilities.

Key takeaways

• Bernoulli has two outcomes: $Y=1$ (success) with $p$, $Y=0$ (failure) with $q=1-p$.
• $E[Y] = p$, $\text{Var}(Y) = p q$ – derived directly from the definitions.
• It is the foundation for the Binomial distribution (multiple independent Bernoulli trials).

Binomial Random Variables and its Distribution

The binomial random variable counts the number of successes in a fixed number of independent Bernoulli trials. Intuitively: if you flip a coin 10 times, how many heads do you get? That count is a binomial random variable.

Binomial Experiment

A binomial experiment has four properties:

1. A sequence of $n$ identical and independent trials.
2. Each trial has exactly two outcomes: success (probability $p$) and failure (probability $q = 1-p$).
3. The probability of success $p$ is constant across all trials.
4. The random variable $Y$ = number of successes in $n$ trials.

Exam tip: The binomial random variable only counts successes – it ignores the order in which they occur. All arrangements of $y$ successes and $n-y$ failures are grouped into the single outcome $Y=y$.

Probability Mass Function (PMF)

The PMF of $Y$ gives the probability of exactly $y$ successes:

$P(Y = y) = \binom{n}{y} \, p^{\,y} \, (1-p)^{\,n-y}, \quad y = 0,1,\dots,n$

where $\displaystyle \binom{n}{y} = \frac{n!}{y!\,(n-y)!}$ counts the number of ways to arrange $y$ successes among $n$ trials.

Why this formula works

• $p^{\,y}$ : probability of $y$ successes.
• $(1-p)^{\,n-y}$ : probability of $n-y$ failures.
• $\binom{n}{y}$ : number of distinct sequences containing exactly $y$ successes.

Examples

Expectation and Variance

• Expectation: $E[Y] = n\,p$
• Variance: $\text{Var}(Y) = n\,p\,q = n\,p\,(1-p)$
• Standard deviation: $\sqrt{n\,p\,(1-p)}$

Intuition: The binomial random variable is the sum of $n$ independent Bernoulli($p$) random variables (each with mean $p$ and variance $p(1-p)$). Hence the mean and variance simply add up.

Worked example (fair die)

For $n=15$, $p=1/3$:

$E[Y] = 15 \times \frac{1}{3} = 5$ $\text{Var}(Y) = 15 \times \frac{1}{3} \times \frac{2}{3} = \frac{10}{3} \approx 3.33$ $\text{SD}(Y) = \sqrt{\frac{10}{3}} \approx 1.83$

This means over many sets of 15 rolls, you would expect about 5 successes (rolling ≤2), give or take about 1.8.

Key takeaways

• A binomial random variable requires: fixed $n$, independent trials, two outcomes, constant $p$.
• PMF: $\displaystyle P(Y=y) = \binom{n}{y} p^y (1-p)^{n-y}$.
• Mean $= np$; variance $= np(1-p)$.
• The binomial counts successes – it does not track which trials succeeded.
• Common pitfalls: using binomial when trials are not independent (e.g., sampling without replacement) – use hypergeometric instead.

Exam tip: The binomial PMF sums to 1: $\sum_{y=0}^n \binom{n}{y} p^y (1-p)^{n-y} = 1$. This is the binomial theorem in action. Memorise the mean and variance formulas – they appear in nearly every exam problem.

Examples of Binomial Distributions

The binomial distribution models the number of successes in a fixed number of independent trials, each with the same success probability $p$. The examples that follow show how to compute probabilities, expectations, and perform inverse calculations using a spreadsheet.

Key formulas (for binomial random variable $Y \sim \text{Bin}(n,p)$)

• PMF: $P(Y=y) = \binom{n}{y} p^y (1-p)^{n-y}$
• Expected value: $\mu = E[Y] = np$
• Variance: $\sigma^2 = np(1-p)$
• Standard deviation: $\sigma = \sqrt{np(1-p)}$

Excel: BINOM.DIST function

A template table lists $y$ from $0$ to $n$, then computes $P(Y=y)$ (PMF), $P(Y \le y)$ (CDF), and $P(Y \ge y)$ (complement) for each $y$. The spreadsheet also automatically updates the mean, variance, and standard deviation when $n$ or $p$ changes.

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Example 1: Biased die (10 tosses, success = 5 or more)

• Trials: $n = 10$
• Success: outcome $\ge 5$ on a die $\Rightarrow p = 1/6$
• Failure: outcome $\le 4$ $\Rightarrow q = 5/6$
• Random variable: $Y$ = number of successes in 10 tosses

Computed parameters

• Mean: $E[Y] = np = 10 \times \frac{1}{6} \approx 1.67$
• Variance: $npq = 10 \times \frac{1}{6} \times \frac{5}{6} \approx 1.39$
• Standard deviation: $\sqrt{1.39} \approx 1.18$

Probability of at least 3 successes

$P(Y \ge 3) = \sum_{y=3}^{10} P(Y=y)$

From the spreadsheet, summing the PMF values for $y=3,4,\dots,10$ gives:

$P(Y \ge 3) \approx 0.225$

Inverse probability: find $p$ such that $P(Y \ge 3) \ge 0.5$

Starting from $p=1/6$, the probability $P(Y \ge 3)$ is $0.225$. By incrementally increasing $p$ in the spreadsheet cell, the desired probability reaches $0.5$ when $p \approx 0.259$.

• $p = 0.18 \rightarrow P(Y \ge 3) \approx 0.263$
• $p = 0.20 \rightarrow 0.322$
• $p = 0.24 \rightarrow 0.44$
• $p = 0.25 \rightarrow 0.474$
• $p = 0.259 \rightarrow 0.501$

The goal seek feature in Excel automates this.

Exam tip: An inverse probability problem gives you a target probability and asks for the parameter $p$ (or sometimes $n$) that produces it. The spreadsheet approach (trial and error or goal seek) is efficient but on an exam you may need to reason qualitatively or use a calculator.

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Example 2: Kedar Apte’s pump defects (14 days, success = day with ≤1 defect)

• Trials: $n = 14$ independent days
• Success: day with one or fewer defective pumps $\Rightarrow p = 0.65$
• Failure: day with two or more defects $\Rightarrow q = 0.35$
• Random variable: $Y$ = number of success days in 14 days

Computed parameters

• Mean: $E[Y] = 14 \times 0.65 = 9.1$
• Standard deviation: $\sqrt{14 \times 0.65 \times 0.35} \approx 1.78$

Probability of at least 10 success days

$P(Y \ge 10) = \sum_{y=10}^{14} P(Y=y)$

From the spreadsheet:

$P(Y \ge 10) \approx 0.423$

Inverse probability: find $p$ such that $P(Y \ge 10) \ge 0.5$

Current $p=0.65$ gives $P(Y \ge 10)=0.423$. Increase $p$:

• $p = 0.66 \rightarrow 0.454$
• $p = 0.67 \rightarrow 0.486$
• $p = 0.68 \rightarrow 0.519$

Thus a success probability of approximately $0.675$ (between 0.67 and 0.68) yields $P(Y \ge 10)$ just over $0.5$.

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How the spreadsheet approach works (diagram)

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Key takeaways

• The binomial distribution is fully determined by $n$ and $p$; all probabilities, mean, and variance follow.
• Spreadsheet templates (with BINOM.DIST) simplify repeated calculations and allow quick parameter sensitivity analysis.
• An inverse probability problem fixes a tail probability and solves for $p$ (or $n$) – often by trial and error or goal seek.
• Both examples illustrate the same workflow: define success → set $n$ and $p$ → compute probabilities → answer questions about expectation, standard deviation, and tail events.

Binomial Distribution – Examples

A binomial random variable $Y$ counts the number of successes in $n$ independent identical Bernoulli trials, each with success probability $p$ (fail probability $q=1-p$).

• PMF: $\displaystyle P(Y=y) = \binom{n}{y} p^y q^{n-y}, \quad y=0,1,\dots,n$
• Mean: $\mu = np$
• Variance: $\sigma^2 = npq$

In practice, cumulative probabilities are obtained from a spreadsheet or binomial tables.

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Example 1 – Mangesh Nadkarni (Plane Delays)

• Success = “day with ≤2 delays” → $p=0.6$, $q=0.4$.
• Trials: $n=30$ independent days.
• $Y$ = # of success days.

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Example 2 – Priya (Insurance Calls)

• Success = “call leads to purchase” → $p=0.2$, $q=0.8$.
• Trials: $n=12$ calls per day.
• $Y$ = # of successes.

Expected number of successes: $\mu = np = 12 \times 0.2 = 2.4$. Standard deviation: $\sigma = \sqrt{npq} = \sqrt{12 \times 0.2 \times 0.8} \approx 1.39$.

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Example 3 – Baburao (Star Days)

• Success = “sales >10 000” → $p=0.3$, $q=0.7$.
• Trials: $n=10$ days.
• $Y$ = # of star days.

Expected star days: $\mu = np = 10 \times 0.3 = 3$.

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Redefining Success – The Critical Pitfall

The same problem context can require different definitions of “success”. Always re‑identify $p$ and $n$ for each question.

Example – Kedar Apte (Defective Pumps)

Original success: “≤1 defect per day” → $p=0.65$. But new question: “at least 7 days with no defects” → success = “0 defects” → $p=0.4$ (from the distribution).

• $n=14$, $P(Y \ge 7)=0.308$.

Another question: “at most 3 days with ≥3 defects” → success = “≥3 defects” → $p=0.15$.

• $n=7$, $P(Y \le 3)=0.988$.

Yet another: “at most one defect on all 14 days” → success = “≤1 defect” ($p=0.65$), but event is $Y=14$.

• $P(Y=14)=0.002$.

Exam tip: Never carry forward a previous $p$ without rechecking what “success” means in the current question. The binomial calculation is mechanical; the hard part is mapping the business question to the correct $n$ and $p$.

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More Complex Examples – Mangesh Nadkarni (Plane Delays Revisited)

Based on historical data for late planes, with probabilities treated as known:

The probability of ≥15 successes with $p=0.3$ is low; the manager is optimistic.

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Divide-and-Conquer Procedure

Key takeaways

• Binomial distribution models counts of successes in a fixed number of independent Bernoulli trials with constant $p$.
• The formulas $P(Y=y)=\binom{n}{y}p^y q^{n-y}$, $\mu=np$, $\sigma^2=npq$ are the core tools.
• The most common errors are misidentifying $p$ (success definition) and $n$ (number of trials). Always start by clarifying these.
• Use complement or subtraction tricks (e.g., $P(Y\le a) = 1-P(Y\ge a+1)$) when helpful, but spreadsheet sums are straightforward.
• Cumulative probabilities (at most, at least, between) are the most frequent exam questions – practice translating business language into probability events.

Recap of Binomial Distribution

A binomial experiment consists of a sequence of identical and independent trials. Each trial has exactly two outcomes — success (probability $p$) and failure (probability $q = 1-p$). The probability $p$ remains constant across trials. The binomial random variable $Y$ counts the number of successes in $n$ such trials. Equivalently, it is the sum of $n$ identical and independently distributed (IID) Bernoulli random variables.

If any assumption fails — e.g., success probabilities change or trials are dependent — the experiment is no longer binomial. In practice, other discrete distributions are then used to model the situation.

Key takeaways

• Binomial: fixed $n$, constant $p$, independence, two outcomes per trial.
• $Y \sim \text{Binomial}(n,p)$ with PMF $P(Y=y) = \binom{n}{y} p^y q^{n-y}$.
• $E[Y] = np$, $Var[Y] = npq$.
• When assumptions break, alternative distributions are needed.

Poisson Distribution

The Poisson random variable counts the number of occurrences of an event over a specified interval of time or space. It is a discrete distribution (values $0,1,2,\dots$) and is often used to model random arrivals, defects, or counts that occur at a constant average rate.

Intuition and Examples

• Number of customers arriving at a store in 30 minutes.
• Number of phone calls at a call center in 15 minutes.
• Number of defects in one kilometre of highway.
• Number of keyword occurrences on a page of text.

Probability Mass Function (PMF)

Let $Y$ be Poisson with parameter $\Lambda > 0$ (the mean number of events in the interval). The probability of exactly $y$ events is:

$P(Y=y) = \frac{e^{-\Lambda}\,\Lambda^y}{y!}, \quad y = 0,1,2,\dots$

$e \approx 2.71828$ is the base of the natural logarithm.

Exam tip: Poisson has only one parameter $\Lambda$, unlike binomial’s two ($n,p$).

Expectation and Variance

For a Poisson random variable $Y$:

$E[Y] = \Lambda, \qquad Var[Y] = \Lambda$

Thus the mean equals the variance — a unique property of the Poisson distribution. The standard deviation is $\sqrt{\Lambda}$.

Properties of a Poisson Process

A Poisson process is a stochastic process where events occur randomly over time, and the number of events in any interval of length $t$ follows a Poisson distribution. The process satisfies:

1. Stationarity – The probability of a given number of events in an interval depends only on the interval’s length, not its start time.
2. Independence – The occurrence or non‑occurrence of events in disjoint intervals are independent.
3. Mean equals variance – The number of events in any interval has identical mean and variance.

The time between successive events in a Poisson process follows an exponential distribution (a continuous distribution covered later).

Parameter as a Rate

When interest lies in the number of events over an interval of length $t$, the Poisson parameter becomes $\Lambda = \lambda t$, where $\lambda$ is the rate (events per unit time). Then $Y$ – the number of events in time $t$ – is Poisson with mean $\lambda t$:

$P(Y=y) = \frac{e^{-\lambda t}\,(\lambda t)^y}{y!}$

Worked Example: Call Centre Arrivals

Suppose customers call a help desk at a rate of $\lambda = 10$ per hour.

• In a $t = 2$‑hour period, the number of calls $Y$ is Poisson with $\Lambda = 10 \times 2 = 20$. $E[Y] = 20$, $Var[Y] = 20$.
• In a $t = 30$‑minute ($0.5$ hour) period, the number of calls $Z$ is Poisson with $\Lambda = 10 \times 0.5 = 5$. $E[Z] = 5$, $Var[Z] = 5$.

The PMF in each case uses the corresponding $\Lambda$.

Connection to the Mathematician

The distribution is named after Baron Siméon Denis Poisson (1781–1840), a French mathematician and physicist.

Key takeaways

• Poisson models counts of rare events over time or space.
• PMF: $P(Y=y) = e^{-\Lambda}\Lambda^y / y!$, with single parameter $\Lambda$.
• $E[Y] = Var[Y] = \Lambda$.
• For a process with rate $\lambda$ over interval $t$, parameter $\Lambda = \lambda t$.
• Key properties: stationarity, independence, mean=variance.

Poisson Distribution: Examples and Excel Implementation

The Poisson distribution models the number of events occurring in a fixed interval of time (or space) when events happen at a constant average rate and independently. The distribution has a single parameter $\lambda$ (also denoted as the mean and variance). The probability mass function (PMF) is:

$P(Y = y) = \frac{e^{-\lambda} \lambda^y}{y!}, \quad y = 0, 1, 2, \dots$

Setting Up Poisson in Excel

Excel does not have a dedicated Poisson function, but the PMF can be built using standard functions:

This table allows quick computation and plotting of the PMF for any $\lambda$.

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Example 1: Customer Arrivals at an Eatery

At Sheetal Dadava, the number of customers arriving in one hour follows a Poisson distribution with $\lambda = 10$. Questions and solutions:

*$\lambda$ scales with time: for a 30‑minute interval, $\lambda = 10 \times \frac{30}{60} = 5$ (the process is Poisson; the rate is proportional to interval length).

Exam tip: When the time interval changes, always rescale $\lambda$. “More than 10” means $Y > 10$, which equals $Y \ge 11$. Reading the wrong row is a common trap.

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Example 2: Social Media Likes

Sriram’s weekly likes follow a Poisson distribution with $\lambda = 14$. Similarly, daily likes use $\lambda = 14 / 7 = 2$.

Exam tip: For “between … and …” always check inclusiveness. If both endpoints are included, sum the PMF values directly. The tail probability column gives $P(Y \ge y)$, not $P(Y > y)$.

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Key Takeaways – Poisson Examples

• The Poisson PMF is $P(Y=y) = e^{-\lambda}\lambda^y / y!$; $\lambda$ = mean = variance.
• Excel setup (no built‑in function) uses EXP, ^, and FACT to compute PMF, then cumulative and tail probabilities by iteration.
• When the time interval changes, $\lambda$ scales proportionally (e.g., half the time → half the $\lambda$).
• Translating a business question into a probability statement is critical: “more than 10” = $Y > 10 = Y \ge 11$; “at most 10” = $Y \le 10$; “at least 10” = $Y \ge 10$.
• For range probabilities (between a and b inclusive), sum the PMF for each $y$ in that range.

Examples of Poisson Distribution (Call Center)

The Poisson distribution models the number of events (e.g., phone calls) occurring in a fixed interval of time when events happen independently at a constant average rate. Its only parameter is the mean $\lambda$ (the average number of events per interval). The probability mass function is

$P(Y = y) = \frac{e^{-\lambda} \lambda^y}{y!}, \quad y = 0,1,2,\dots$

A classic application is call‑center staffing. A call center receives calls according to a Poisson process: the number of calls in any interval depends only on the interval length, not on the time of day, and successive intervals are independent. (More advanced models use non‑stationary Poisson processes with time‑varying rates.)

Scaling the Poisson Parameter $\lambda$

The average rate is given for one interval but may be needed for another. If the rate is constant, $\lambda$ scales linearly with the interval length.

Example: The call center u.net averages 6 calls per 15 minutes. Rate per minute = $6 / 15 = 0.4$ calls/min. For any $t$‑minute interval, $\lambda = 0.4 \cdot t$.

Worked Calculations

1. Probability of 3 calls in 5 minutes

$\lambda = 2$, $y=3$:

$P(Y=3) = \frac{e^{-2} 2^3}{3!} = \frac{0.1353 \times 8}{6} \approx 0.1804$

There is roughly an 18% chance of exactly 3 calls in a 5‑minute window.

2. Probability of 10 calls in 15 minutes

$\lambda = 6$, $y=10$:

$P(Y=10) = \frac{e^{-6} 6^{10}}{10!} \approx 0.0413$

A 4.1% chance – a relatively low probability, suggesting such high demand may strain staffing.

3. Probability of no calls during a 3‑minute break

$\lambda = 1.2$, $y=0$:

$P(Y=0) = e^{-1.2} \approx 0.3012$

Only a 30% chance that a 3‑minute break will be uninterrupted. The agent is likely to be called back before the break ends.

Key Takeaways

• The Poisson distribution depends entirely on the mean $\lambda$ for the given interval.
• Scaling: $\lambda$ for a new interval = (original rate per unit time) × (new interval length).
• Probabilities are computed directly from the PMF (or a spreadsheet).
• Call‑center applications illustrate how randomness affects service quality and staffing decisions.

Exam tip: Always check that the $\lambda$ you use matches the interval of the question. A common mistake is to use the original $\lambda$ without rescaling.

Poisson Distribution – Worked Examples

The Poisson distribution models the count of events occurring in a fixed interval of time or space when events happen independently at a constant average rate. The single parameter $\lambda$ (often called the rate parameter) equals both the mean and variance:

$P(Y = y) = \frac{e^{-\lambda} \lambda^y}{y!}, \quad y = 0,1,2,\dots \qquad \mathbb{E}[Y] = \lambda,\quad \operatorname{Var}(Y) = \lambda$

When the interval changes, $\lambda$ scales proportionally.

Example 1: Mall Visitor Arrivals

Context: Prozone Mall, Aurangabad. Average visits during busy hour = 180 per hour. Manager Mansoor wants the probability of 10–20 people accumulating in a 10‑minute window.

Adjusting $\lambda$ for the new interval

• 180 per hour $\rightarrow$ 10 minutes: $\lambda = 180 \times \frac{10}{60} = 30$ visitors per 10 min
• Similarly, if average is 150/hr → $\lambda = 25$; if 210/hr → $\lambda = 35$

Probabilities (from Poisson PMF)

Why does the probability decrease as $\lambda$ increases? The Poisson distribution becomes roughly symmetric around its mean. As $\lambda$ shifts right (25 → 30 → 35), the fixed window [10,20] captures less probability mass because the distribution moves to higher counts.

Exam tip: Always re‑scale $\lambda$ to match the interval of interest. The probability over the same numeric range can rise or fall as $\lambda$ changes – your intuition about “more arrivals = more in the window” fails when the window is far from the mean.

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Example 2: Highway Defects

Context: Maharashtra State Highway No. 3. Average defects = 3 per km. Count of defects in a stretch follows Poisson.

Probabilities for 1 km stretch ($\lambda=3$)

• $P(Y \ge 3) = 0.577$ (tail probability)
• $P(Y \le 5) = 0.916$ (cumulative)

Probabilities for an 8 km stretch ($\lambda = 3 \times 8 = 24$)

• $P(Y \le 40) = 0.999$
• $P(Y \ge 20) = 0.82$

Common mistake: Trying to compute $P(Y \le 40 \text{ in 8 km})$ as $[P(Y \le 5 \text{ in 1 km})]^8$ (or $0.916^5$ as initially attempted). This is wrong because it assumes every 1‑km segment must have ≤5 defects – the total can be ≤40 even if some segments exceed 5, as long as others compensate.

Exam tip: For a Poisson process, the total count over combined independent intervals is Poisson with $\lambda_{\text{total}} = \lambda_{\text{unit}} \times \text{number of units}$. NEVER multiply probabilities of sub‑intervals.

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Example 3: Social Media Flags

Context: NIA analyst Nitin monitors flags per conversation. Average = 5 flags per conversation. Flags occur independently and at constant rate.

Single conversation ($\lambda=5$)

• $P(2 \le Y \le 8) = 0.891$
• $P(Y \ge 10) = 0.032$ (less than 3% – low alert)

Two consecutive conversations ($\lambda = 5+5 = 10$)

• $P(4 \le Y \le 16) = 0.963$
• $P(Y < 4) = P(Y \le 3) = 0.01$ (very unlikely)

Note that “less than 4 flags in two conversations” means $Y \le 3$, not $Y \le 4$.

Result summary table

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Key Properties of the Poisson Distribution

• Constant rate – probability of an event is the same in any two intervals of equal length.
• Independence – occurrence in one interval is independent of occurrence in any other non‑overlapping interval.
• Mean = Variance = $\lambda$.

These properties make Poisson suitable for counts over time (arrivals, calls, flags) and over space (defects per km, potholes, errors in wafers or text).

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Connection Between Binomial and Poisson

When $p$ (success probability) is small and $n$ is large, the binomial distribution $\operatorname{Bin}(n,p)$ approximates a Poisson with $\lambda = np$.

• Binomial: $\mathbb{E}[Y]=np$, $\operatorname{Var}(Y)=np(1-p)$
• Poisson: $\mathbb{E}[Y]=\operatorname{Var}(Y)=\lambda$

If $p$ is small, $1-p \approx 1$, so binomial variance ≈ $np = \lambda$, matching Poisson. Many Poisson distributions also appear symmetric and bell‑shaped for moderate $\lambda$ – a preview of the Central Limit Theorem.

Exam tip: Use the Poisson approximation to the binomial when $n$ is large, $p$ is small, and $np$ is moderate (rule of thumb: $n \ge 20$, $p \le 0.05$).

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Key Takeaways

• Poisson models counts of rare, independent events over time/space; one parameter $\lambda$ = mean = variance.
• When interval changes, $\lambda$ scales proportionally; always recompute for the exact interval.
• Probabilities computed via PMF, cumulative (CDF), or tail – spreadsheet tools are helpful.
• Common pitfalls: multiplying probabilities across intervals instead of summing means; confusing $P(Y < k)$ with $P(Y \le k)$.
• The binomial approximates Poisson when $p$ is small ($\lambda = np$).