L5.1 Discrete Distributions: Bernoulli and Binomial Distributions

Discrete Probability Distributions

Introduction to Discrete Distributions

Many real-world problems in business, economics, and other fields involve discrete random variables, which represent outcomes that can only take on specific, separate values (such as counts or binary outcomes). To model the randomness of such variables, we use specific probability distributions. Two important and widely used discrete distributions are the Binomial distribution and the Poisson distribution.

• Binomial distribution: Used for modeling the number of successes in a fixed number of independent trials, such as click-through rates in online advertising, disease detection (positive/negative), or the number of defective items in quality control.

• Poisson distribution: Used for modeling the number of events occurring in a fixed interval of time or space, such as the number of purchases or calls per customer, product failures over time, or the number of accidents per day.

Bernoulli Distribution

Definition and Properties

The Bernoulli distribution is the simplest discrete probability distribution, modeling a random variable that takes the value 1 ("success") with probability p and the value 0 ("failure") with probability 1 − p, where 0 < p < 1.

• Notation: $X \sim \mathrm{Ber}(p)$

x	0	1
p(x)	1-p	p

The probability mass function (PMF) in function form:

• $p(x) = \begin{cases} 1-p & \text{if } x=0, \\ p & \text{if } x=1, \\ 0 & \text{otherwise} \end{cases}$

• Alternatively: $p(x) = p^x (1-p)^{1-x}$ for $x=0$ or $1$

Expected Value and Variance

• Expected value: $E[X] = \sum_x x p(x) = 0 \times (1-p) + 1 \times p = p$

• Variance: $E[X^2] = \sum_x x^2 p(x) = 0^2 \times (1-p) + 1^2 \times p = p$ $\mathrm{Var}(X) = E[X^2] - (E[X])^2 = p - p^2 = p(1-p)$

Example: Bernoulli Distribution in Quality Control

A shipment of DVD players contains three defective and seven nondefective players. If a customer randomly selects a player, let $x=1$ if the player is defective and $x=0$ otherwise. The probability of selecting a defective player is $p=0.3$, so $X \sim \mathrm{Ber}(0.3)$.

Binomial Distribution

Definition and Properties

The Binomial distribution models the number of successes in a fixed number of independent Bernoulli trials, each with the same probability of success p.

• Notation: $X \sim \mathrm{Bin}(n, p)$

• $X$ can take values $x = 0, 1, \ldots, n$

Probability Mass Function (PMF)

• $p(x) = \begin{cases} {n \choose x} p^x (1-p)^{n-x} & \text{if } x = 0, 1, \ldots, n \\ 0 & \text{otherwise} \end{cases}$

• The term ${n \choose x} = \frac{n!}{x!(n-x)!}$ counts the number of ways to choose $x$ successes in $n$ trials.

• The term $p^x (1-p)^{n-x}$ is the probability of having exactly $x$ successes and $n-x$ failures.

Example: Coin Flips

Suppose we flip three coins, each showing heads with probability 0.2. Let $X$ denote the number of heads obtained. Then $X \sim \mathrm{Bin}(3, 0.2)$.

• $P(X=0) = P(\{TTT\}) = 0.8^3 = 0.512$

• $P(X=1) = P(\{TTH, THT, HTT\}) = 3 \times 0.2^1 \times 0.8^2 = 0.384$

• $P(X=2) = P(\{THH, HTH, HHT\}) = 3 \times 0.2^2 \times 0.8^1 = 0.096$

• $P(X=3) = P(\{HHH\}) = 0.2^3 = 0.008$

Relationship to Bernoulli Distribution

• If $Y_i$, $i=1,\ldots,n$ are independent and identically distributed Bernoulli random variables with parameter $p$, then $X = Y_1 + Y_2 + \cdots + Y_n \sim \mathrm{Bin}(n, p)$.

• The Bernoulli distribution is a special case of the binomial distribution with $n=1$.

Expected Value and Variance

• Expected value: $E[X] = np$

• Variance: $\mathrm{Var}(X) = np(1-p)$

Example: Payment Method in a Store

Suppose 30% of customers pay by credit card. For the next five customers, what is the probability that three pay by credit card?

• Let $X$ be the number of customers paying by credit card: $X \sim \mathrm{Bin}(5, 0.3)$

• $P(X=3) = {5 \choose 3} (0.3)^3 (0.7)^2 = 10 \times 0.027 \times 0.49 = 0.1323$

Example: Defective Light Bulbs

A factory produces light bulbs, each with a 5% chance of being defective. If 20 bulbs are tested, let $X$ be the number of defective bulbs.

• $X \sim \mathrm{Bin}(20, 0.05)$

• Probability exactly 2 are defective: $P(X=2) = {20 \choose 2} (0.05)^2 (0.95)^{18} \approx 0.1887$

• Expected number: $E[X] = 20 \times 0.05 = 1$

• Variance: $\mathrm{Var}(X) = 20 \times 0.05 \times 0.95 = 0.95$

• Probability exactly 2 are defective, given at least one is defective: $P(X=2 \mid X \geq 1) = \frac{P(X=2)}{1 - P(X=0)} = \frac{0.1887}{1 - (0.95)^{20}} \approx 0.2941$

Summary Table: Bernoulli vs. Binomial Distribution

Property	Bernoulli	Binomial
Number of trials	1	n
Possible values	0, 1	0, 1, ..., n
PMF	$p^x (1-p)^{1-x}$	${n \choose x} p^x (1-p)^{n-x}$
Mean	$p$	$np$
Variance	$p(1-p)$	$np(1-p)$

References: Mittelhammer, R. C. (2013). Mathematical Statistics for Economics and Business. Springer.