MathIsimple

Lesson 5-3: Probability Models & Expected Value

Focus

  • Compute expected value and variance for discrete models.
  • Apply binomial and Poisson distributions to counts.
  • Use EV to guide decisions under uncertainty.

Prerequisites

  • Basic probability rules and combinatorics.
  • Comfort with summations and interpreting parameters.

Binomial Distribution

Definition

X∼Binomial(n,p)X \sim \text{Binomial}(n,p) models the number of successes in n independent trials with success probability p.

P(X=k)=(nk)pk(1−p)n−kP(X=k)=\binom{n}{k} p^k (1-p)^{n-k}
E[X]=np,  Var(X)=np(1−p)E[X]=np, \; Var(X)=np(1-p)

Applications

  • Quality control: count defects in a sample of size n.
  • Marketing: number of conversions in n outreach attempts.
  • Reliability: successes before failure thresholds.

Poisson Distribution

Definition

X∼Poisson(λ)X \sim \text{Poisson}(\lambda) models counts of rare, independent events in a fixed interval with rateλ\lambda.

P(X=k)=e−λλkk!P(X=k)=e^{-\lambda} \dfrac{ \lambda^k }{ k! }
E[X]=λ,  Var(X)=λE[X]=\lambda, \; Var(X)=\lambda

Applications

  • Call arrivals in a contact center per minute.
  • Defects per meter of material on a production line.
  • Requests per second to a web service.

Expected Value in Decisions

Expected value summarizes long-run average outcomes. In planning, compare EVs across choices and consider variance to assess risk.

Lottery Example

Payouts: 5,000 (0.4), 1,000 (0.5), -2,000 (0.1).

E[X]=5000⋅0.4+1000⋅0.5−2000⋅0.1=2300E[X]=5000\cdot0.4 + 1000\cdot0.5 - 2000\cdot0.1 = 2300

Quality Planning

With defect probability p and batch size N, expected defects are NpNp; budget sampling accordingly.

Worked Example: Binomial

Shooter accuracy p=0.8p=0.8, 5 shots. Compute P(X=3)P(X=3), E[X]E[X], andVar(X)Var(X).

P(X=3)=(53)0.830.22=10×0.512×0.04=0.2048P(X=3)= \binom{5}{3} 0.8^3 0.2^2 = 10\times0.512\times0.04=0.2048
E[X]=np=4,  Var(X)=np(1−p)=0.8E[X]=np=4, \; Var(X)=np(1-p)=0.8

Worked Example: Poisson

A machine averages λ=2\lambda=2 breakdowns per month. Compute P(X=0)P(X=0) andP(X≥3)P(X\ge3) for one month.

P(X=0)=e−2200!=e−2P(X=0)=e^{-2} \dfrac{2^0}{0!}=e^{-2}
P(X≥3)=1−[e−2200!+e−2211!+e−2222!]P(X\ge3)=1-\left[ e^{-2} \dfrac{2^0}{0!}+e^{-2} \dfrac{2^1}{1!}+e^{-2} \dfrac{2^2}{2!} \right]

Practice & Projects

Set A: Binomial Quality Control

  1. n, p from context
  2. Compute P(X≤k)
  3. Decide accept/reject lot

Set B: Poisson Web Requests

  1. λ per second
  2. Compute P(X≥k) overload risk
  3. Provision capacity

Set C: Insurance EV

  1. Outcomes and probs
  2. Compute EV and Var
  3. Price with risk loading

Set D: Marketing Conversions

  1. Binomial with n outreach
  2. EV conversions
  3. Budget decision

Set E: Maintenance Scheduling

  1. Poisson failures λ
  2. Expected downtime
  3. Plan spare parts

Set F: Reliability Target

  1. Binomial pass rate
  2. Find p s.t. P(X≥m)≥target
  3. Interpret tolerance

Variance, Risk, and Decision

Compare choices by EV and variance; risk-averse contexts prefer smaller variance given similar EV.

Practice Bank

Bank A: Model Choice

  1. Binomial vs Poisson
  2. Check independence/rare-event conditions
  3. Parameter estimation

Bank B: EV & Variance

  1. Compute E[X], Var(X)
  2. Coefficient of variation
  3. Risk comparison

Bank C: Tail Probabilities

  1. P(X≥k) and P(X≤k)
  2. Normal approximation where valid
  3. Continuity correction note

Bank D: Decision Rules

  1. Threshold policy
  2. Expected cost minimization
  3. Sensitivity to parameters

FAQ (Extended)

Q: EV is higher but variance is large — how to choose?

Depends on risk preference; consider expected utility or add risk penalties.

Q: When to use normal approximation for binomial?

When np and n(1−p) are sufficiently large (e.g., ≥ 10); check continuity corrections.