Discrete Probability Distributions

What is a Discrete Distribution?

A discrete probability distribution describes the probabilities for a random variable that can only take on countable values (integers, specific outcomes, etc.).

Examples of discrete variables:

Number of heads in 10 coin flips (0, 1, 2, …, 10)
Number of customers per hour (0, 1, 2, 3, …)
Number of defective items in a batch (0, 1, 2, …)
Dice roll outcome (1, 2, 3, 4, 5, 6)

Probability Mass Function (PMF)

The probability mass function P(X = x) gives the probability that X equals each possible value.

PMF Properties

$P(X = x) \geq 0$ for all values
$\sum_{\text{all } x} P(X = x) = 1$

Expected Value and Variance

Expected Value (Mean)

$E(X) = \mu = \sum_{x} x \cdot P(X = x)$

Variance

$Var(X) = \sigma^2 = \sum_{x} (x - \mu)^2 \cdot P(X = x) = E(X^2) - [E(X)]^2$

1. Binomial Distribution

The binomial distribution models the number of successes in n independent trials, each with probability p of success.

Conditions for Binomial

Fixed number of trials (n)
Two outcomes per trial (success/failure)
Independent trials
Same probability (p) for each trial

Binomial Distribution

$X \sim \text{Binomial}(n, p)$

$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$

Where:

n = number of trials
k = number of successes
p = probability of success on each trial
$\binom{n}{k}$ = “n choose k” = $\frac{n!}{k!(n-k)!}$

Binomial Mean and Variance

$\mu = np$ $\sigma^2 = np(1-p)$ $\sigma = \sqrt{np(1-p)}$

Free Throw Shooting

A basketball player makes 80% of free throws. In 10 attempts:

P(exactly 8 makes)? $P(X=8) = \binom{10}{8}(0.8)^8(0.2)^2$ $= 45 \times 0.1678 \times 0.04 = 0.302$

P(at least 8 makes)? $P(X \geq 8) = P(8) + P(9) + P(10)$ $= 0.302 + 0.268 + 0.107 = 0.677$

Expected makes: $\mu = 10 \times 0.8 = 8$

Standard deviation: $\sigma = \sqrt{10 \times 0.8 \times 0.2} = \sqrt{1.6} \approx 1.26$

Quality Control

A batch has 5% defective items. Randomly select 20 items.

P(exactly 2 defective)? $P(X=2) = \binom{20}{2}(0.05)^2(0.95)^{18}$ $= 190 \times 0.0025 \times 0.397 = 0.189$

P(at most 1 defective)? $P(X \leq 1) = P(0) + P(1)$ $= \binom{20}{0}(0.05)^0(0.95)^{20} + \binom{20}{1}(0.05)^1(0.95)^{19}$ $= 0.358 + 0.377 = 0.735$

About 73.5% chance of at most 1 defective item.

2. Poisson Distribution

The Poisson distribution models the count of rare events in a fixed interval of time or space, when events occur independently at a constant average rate.

Conditions for Poisson

Events occur independently
Events occur at a constant average rate (λ)
Two events can’t occur at exactly the same instant
We’re counting events in a fixed interval

Poisson Distribution

$X \sim \text{Poisson}(\lambda)$

$P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}$

Where:

λ (lambda) = average number of events per interval
k = number of events (0, 1, 2, …)
e ≈ 2.71828

Poisson Mean and Variance

$\mu = \lambda$ $\sigma^2 = \lambda$ $\sigma = \sqrt{\lambda}$

Call Center

A call center receives an average of 3 calls per minute.

P(exactly 5 calls in a minute)? $P(X=5) = \frac{3^5 e^{-3}}{5!} = \frac{243 \times 0.0498}{120} = 0.101$

P(no calls in a minute)? $P(X=0) = \frac{3^0 e^{-3}}{0!} = e^{-3} = 0.0498$

P(at least 1 call)? $P(X \geq 1) = 1 - P(X=0) = 1 - 0.0498 = 0.950$

Changing the Interval

If average is 3 calls per minute, what about 5 minutes?

New λ = 3 × 5 = 15 calls per 5 minutes

P(exactly 12 calls in 5 minutes)? $P(X=12) = \frac{15^{12} e^{-15}}{12!} \approx 0.083$

Binomial vs Poisson

Aspect	Binomial	Poisson
What we count	Successes in n trials	Events in interval
Parameters	n and p	λ
Max possible value	n	Unbounded
Mean	np	λ
Variance	np(1-p)	λ

3. Geometric Distribution

The geometric distribution models the number of trials until the first success.

Geometric Distribution

$X \sim \text{Geometric}(p)$

$P(X = k) = (1-p)^{k-1} p$

Mean: $\mu = \frac{1}{p}$

Variance: $\sigma^2 = \frac{1-p}{p^2}$

Rolling a Six

Roll a fair die until you get a 6.

p = 1/6, so expected rolls until first 6: $\mu = 6$

P(first 6 on the 4th roll)? $P(X=4) = (5/6)^3 (1/6) = 0.579 \times 0.167 = 0.0965$

P(takes more than 6 rolls)? $P(X > 6) = (5/6)^6 = 0.335$

4. Negative Binomial Distribution

The negative binomial models the number of trials until r successes.

Negative Binomial Distribution

$P(X = k) = \binom{k-1}{r-1} p^r (1-p)^{k-r}$

Where k = total trials needed to get r successes.

Mean: $\mu = \frac{r}{p}$

5. Hypergeometric Distribution

The hypergeometric distribution models sampling without replacement from a finite population.

Hypergeometric Distribution

$P(X = k) = \frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}$

Where:

N = population size
K = number of successes in population
n = sample size
k = number of successes in sample

Committee Selection

A committee of 5 is chosen from 8 men and 6 women.

P(exactly 3 women)?

N = 14, K = 6 (women), n = 5, k = 3

$P(X=3) = \frac{\binom{6}{3}\binom{8}{2}}{\binom{14}{5}} = \frac{20 \times 28}{2002} = \frac{560}{2002} = 0.280$

Choosing the Right Distribution

Scenario	Distribution
Fixed n trials, success/failure	Binomial
Count events in interval, rare events	Poisson
Trials until first success	Geometric
Trials until r successes	Negative Binomial
Sampling without replacement	Hypergeometric

Identifying the Distribution

Scenario 1: Number of defective items in 50 sampled from a batch → Binomial (fixed n, success/failure)

Scenario 2: Number of accidents at an intersection per month → Poisson (counting events in time interval)

Scenario 3: Number of interviews until getting a job offer → Geometric (trials until first success)

Scenario 4: Selecting 10 cards from deck, counting hearts → Hypergeometric (without replacement, finite population)

Summary

In this lesson, you learned:

Probability mass function (PMF) gives P(X = x) for discrete variables
Binomial: n independent trials, each with probability p
Poisson: Counting rare events at constant rate λ
Geometric: Trials until first success
Negative binomial: Trials until r successes
Hypergeometric: Sampling without replacement
Each distribution has formulas for mean and variance
Choosing the right distribution depends on the scenario

Practice Problems

1. A coin is flipped 12 times. Find: a) P(exactly 7 heads) b) P(at least 10 heads) c) Expected number of heads

2. Typos occur at a rate of 2 per page. For a single page: a) P(no typos)? b) P(more than 3 typos)?

3. A die is rolled until a 1 appears. a) P(first 1 on the 5th roll)? b) Expected number of rolls?

4. From a bag of 10 red and 15 blue marbles, select 6 without replacement. P(exactly 4 red)?

Click to see answers

1. Binomial(12, 0.5) a) P(X=7) = $\binom{12}{7}(0.5)^{12}$ = 792 × 0.000244 ≈ 0.193 b) P(X≥10) = P(10) + P(11) + P(12) = 0.016 + 0.003 + 0.0002 ≈ 0.019 c) μ = np = 12 × 0.5 = 6

2. Poisson(λ=2) a) P(X=0) = e⁻² ≈ 0.135 b) P(X>3) = 1 - P(X≤3) = 1 - [P(0)+P(1)+P(2)+P(3)] = 1 - [0.135 + 0.271 + 0.271 + 0.180] = 1 - 0.857 ≈ 0.143

3. Geometric(p=1/6) a) P(X=5) = (5/6)⁴(1/6) = 0.482 × 0.167 ≈ 0.080 b) μ = 1/p = 6 rolls

4. Hypergeometric(N=25, K=10, n=6, k=4) $P(X=4) = \frac{\binom{10}{4}\binom{15}{2}}{\binom{25}{6}} = \frac{210 \times 105}{177100} ≈ \textbf{0.124}$

Next Steps

Continue your probability studies:

Continuous Distributions - Normal, exponential, and more
Sampling Distributions - Distribution of sample statistics
Probability Calculator - Practice calculations

What is a Discrete Distribution?

Probability Mass Function (PMF)

Expected Value and Variance

1. Binomial Distribution

Conditions for Binomial

2. Poisson Distribution

Conditions for Poisson

Binomial vs Poisson

3. Geometric Distribution

4. Negative Binomial Distribution

5. Hypergeometric Distribution

Choosing the Right Distribution

Summary

Practice Problems

Next Steps

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