B. Probability Distributions · Discrete Probability Distributions

Solving Problems with Uniform, Binomial and Hypergeometric Distributions

B1.7 B1.4 B1.5About 75 minutes

Learning goals

I can choose the right model (uniform, binomial or hypergeometric) for a problem and justify the choice.
I can solve multi-step problems involving "at least", "at most" and "between" probabilities.
I can solve real-world problems such as hotel cancellations, overbooking and quality control.
I can state the assumptions of my model and say when they might not hold.

Words to know

model assumptions
The conditions a probability model needs (for example, independent trials with the same p for a binomial model).

overbooking
Selling more reservations than there are seats or rooms, expecting some people not to show up.

acceptance sampling
A quality-control method: inspect a random sample from a batch and accept the batch only if the sample has few enough defects.

Which model?

Real problems don't say "use the binomial formula". You have to read the situation and decide. Ask three questions:

Are all the values equally likely? Then the distribution is uniform.
Am I counting successes in a fixed number of trials? If not, none of these three models may fit; consider a simulation.
Are the trials independent with the same probability (with replacement, or a process like coin tosses), or dependent (without replacement from a small group)? Independent: binomial. Dependent: hypergeometric.

Then state the parameters (n and p, or N, r and n) before you calculate.

A decision guide for the three discrete models in this unit. Always check the conditions; if none fits, a simulation may still give an answer.

Worked example

Example 1: Hotel cancellations (from the curriculum)

A hotel finds that 8% of reservations are cancelled, independently. It has 10 reservations for a night. Find the probability distribution of the number of cancellations, and the probability that at least 4 are cancelled.

Model: 10 independent reservations, each cancelled with p = 0.08. X ~ B(10, 0.08).
P(0) = 0.92¹⁰ ≈ 0.4344; P(1) = 10(0.08)(0.92)⁹ ≈ 0.3777; P(2) = 45(0.08)²(0.92)⁸ ≈ 0.1478; P(3) = 120(0.08)³(0.92)⁷ ≈ 0.0343.
P(X ≥ 4) = 1 − [P(0) + P(1) + P(2) + P(3)] ≈ 1 − 0.9942 = 0.0058.
E(X) = 10 × 0.08 = 0.8 cancellations.

P(X ≥ 4) ≈ 0.006, less than 1%. On a typical night the hotel expects fewer than 1 cancellation among 10 reservations.

The cancellations distribution is strongly skewed right. The bars for 4 or more are almost invisible: together they hold only about 0.6% of the probability.

Worked example

Example 2: Airline overbooking

A small airline flight has 20 seats. From experience, each ticket holder shows up with probability 0.9, independently. The airline sells 22 tickets. What is the probability that more passengers show up than there are seats?

X = number who show up, X ~ B(22, 0.9).
Too many passengers means X = 21 or X = 22.
P(21) = C(22, 21)(0.9)²¹(0.1) = 22 × 0.10942 × 0.1 ≈ 0.2407.
P(22) = 0.9²² ≈ 0.0985.
P(X > 20) ≈ 0.2407 + 0.0985 = 0.3392.

About 0.34. Selling 22 tickets for 20 seats would leave someone without a seat on about one flight in three. The airline would have to weigh this against the revenue from empty seats (lesson 12 does this with expected value). Assumption check: people often travel in groups, so show-ups may not really be independent.

Worked example

Example 3: Acceptance sampling

A shipment of 50 bike lights has 3 defective lights. The store inspects 5 lights chosen at random and accepts the shipment only if none of the 5 is defective. Find the probability that the shipment is accepted.

The 5 inspected lights are different lights: sampling without replacement from a small batch. Hypergeometric with N = 50, r = 3, n = 5.
P(X = 0) = C(3, 0)C(47, 5)/C(50, 5) = 1 533 939/2 118 760 ≈ 0.7240.
So P(rejected) ≈ 0.2760.

P(accepted) ≈ 0.724. Even with 3 defective lights (6%), the shipment passes almost three times in four. A store wanting better protection would inspect more lights.

State your assumptions

Strong answers say why a model fits and what could break it. Binomial: are trials really independent? Is p really the same each time? Hypergeometric: is the sample really random? Uniform: are the outcomes really equally likely?

Worked example

Example 4: Uniform, then binomial

Every day a teacher picks a random student number from 1 to 30 to answer the warm-up question. Your number is 17. (a) What is the probability you are picked on a given day? (b) In a 20-day month, what is the probability you are picked at least twice?

(a) Uniform on 1 to 30: P = 1/30.
(b) Each day is an independent trial with p = 1/30. X ~ B(20, 1/30).
P(0) = (29/30)²⁰ ≈ 0.5076. P(1) = 20(1/30)(29/30)¹⁹ ≈ 0.3501.
P(X ≥ 2) = 1 − 0.5076 − 0.3501 ≈ 0.1423.

(a) 1/30; (b) ≈ 0.142. Many problems combine models: a uniform choice each day becomes a binomial count over many days.

Just for fun

A hotel that understands the binomial distribution still leaves the light on for you. It just knows there is an 8% chance you won't need it.

Hands-on activity

Model-choosing card sort

You need: 8 index cards or slips of paper and a pen.

Write one situation on each card:

(a) number of 6s in 12 rolls of a die
(b) number of Grade 12 names drawn when 4 names are pulled from a hat holding 10 Grade 12 and 8 Grade 11 names
(c) the number on a ticket drawn from a raffle drum with tickets 1 to 500
(d) number of defective phones in a sample of 3 from a box of 12 with 2 defective
(e) number of free throws made in 10 tries by an 80% shooter
(f) last digit of a randomly chosen licence plate
(g) number of hearts in 6 cards drawn with replacement
(h) number of rolls needed to get a 6

Sort the cards into four piles: uniform, binomial, hypergeometric, none of these.
For each card, write the parameters (n, p or N, r, n).

Answers to check: binomial: (a) B(12, 1/6), (e) B(10, 0.8), (g) B(6, 1/4). Hypergeometric: (b) N = 18, r = 10, n = 4; (d) N = 12, r = 2, n = 3. Uniform: (c) 1 to 500; (f) 0 to 9 (approximately). None: (h), since the number of trials is not fixed.

Your turn

Name the model and parameters: "A quiz has 12 true-or-false questions. Mo guesses all of them. X = number correct."

Name the model and parameters: "6 cards dealt from a deck; X = number of spades."

Hotel problem: with 10 reservations and p = 0.08, find P(no cancellations).

A survey finds that 60% of a large town supports a new arena. A random sample of 12 residents is taken. Find P(at least 7 support it).

A box has 20 chocolates, 6 with maple filling. You choose 4 at random. Find P(exactly 2 maple).

A raffle drum has tickets numbered 1 to 500. Find P(the winning number is from 451 to 500).

In the acceptance-sampling example (N = 50, r = 3, n = 5), what is the expected number of defective lights in the sample?

More practice: warm-up

Sampling with replacement suggests which model?

Sampling a few items from a small batch without replacement suggests which model?

Random integer from 1 to 12: which model?

X ~ B(10, 0.08). Find E(X).

What does "at least 4" mean for X?

More practice: on your own

About 2% of a model of phone have a faulty battery. A store sells 50. Find P(at most 1 faulty).

A carpool lottery picks 3 names from 15 students, of whom 5 live on the same street. Find P(at least one of the 5 is picked).

An airline sells 22 tickets for 20 seats, p(show) = 0.9. Find P(everyone who shows gets a seat).

A quality inspector will reject a batch of 100 if a sample of 10 contains 2 or more defective items. If 5 items in the batch are defective, which model would you use and why?

For the previous question, find P(rejected) using P(X = 0) ≈ 0.5838 and P(X = 1) ≈ 0.3394.

Challenge

The hotel in Example 1 wants the probability of at least one cancellation to be over 50%. What is the smallest number of reservations n for which this is true (p = 0.08)?

An airline with 20 seats and show probability 0.9 sells 21 tickets. Find P(too many passengers), and compare with selling 22.

A store can choose to inspect 5 or 10 lights from a shipment of 50 with 3 defective, accepting only if no defective light is found. Find P(accept) for each plan.

Check your understanding

Quiz

X = number of red cards when 4 cards are dealt from a deck. Which model fits?

X ~ B(10, 0.08). Which expression gives P(X ≥ 4)?

Which assumption is most likely to fail when modelling airline show-ups as binomial?

A teacher picks a random number from 1 to 25 each day. What is the probability of picking a number from 21 to 25?

A batch of 50 has 3 defective items; a sample of 5 is inspected. What is the expected number of defective items in the sample?

A 60% shooter takes 8 shots. What is the expected number of makes?

For parents and teachers

Big idea (B1.7): the most important skill is choosing the model. Equally likely values → uniform. Counting successes in independent trials with the same p → binomial. Counting successes when sampling without replacement from a small group → hypergeometric. Then state the parameters, calculate, and say what the answer means and what assumptions it relies on.

How to help: do the card sort together. Ask "Why is this one binomial and not hypergeometric?" Good answers mention replacement and independence.

Common difficulties:

Choosing binomial for any problem with "n" and "p" in it, even when sampling is without replacement.
Off-by-one errors in "at least" and "more than".
Forgetting to check whether the answer makes sense (a probability above 1, or an expected value outside the possible range).