# 2.4: Sampling from a Small Population

Example \(\PageIndex{1}\)

Professors sometimes select a student at random to answer a question. If each student has an equal chance of being selected and there are 15 people in your class, what is the chance that she will pick you for the next question?

**Solution**

If there are 15 people to ask and none are skipping class, then the probability is 1=15, or about 0:067.

Example \(\PageIndex{2}\)

If the professor asks 3 questions, what is the probability that you will not be selected? Assume that she will not pick the same person twice in a given lecture.

**Solution**

For the first question, she will pick someone else with probability 14=15. When she asks the second question, she only has 14 people who have not yet been asked. Thus, if you were not picked on the first question, the probability you are again not picked is 13=14. Similarly, the probability you are again not picked on the third question is 12=13, and the probability of not being picked for any of the three questions is

\[ P \text{ (not picked in 3 questions)} = P (Q1 = not-picked, Q2 = not-picked, Q3 = not-picked)\]

\[ = \dfrac {14}{15} \times \dfrac {13}{14} \times \dfrac {12}{13} = \dfrac {12}{15} = 0.80 \]

Exercise \(\PageIndex{1}\)

What rule permitted us to multiply the probabilities in Example \(\PageIndex{2}\)?

**Answer**

The three probabilities we computed were actually one marginal probability, P(Q1=not picked), and two conditional probabilities:

P(Q2 = not picked | Q1 = not picked)

P(Q3 = not picked | Q1 = not picked, Q2 = not picked)

Using the General Multiplication Rule, the product of these three probabilities is the probability of not being picked in 3 questions.

^{42}Each probability is conditioned on the same information that the garage is full, so the complement may be used: \(1.00 - 0.56 - 0.35 = 0.09\).

Example \(\PageIndex{3}\)

Suppose the professor randomly picks without regard to who she already selected, i.e. students can be picked more than once. What is the probability that you will not be picked for any of the three questions?

**Solution**

Each pick is independent, and the probability of not being picked for any individual question is 14=15. Thus, we can use the Multiplication Rule for independent processes.

\[ P \text{ (not picked in 3 questions)} = P (Q1 = not-picked, Q2 = not-picked, Q3 = not-picked)\]

\[ = \dfrac {14}{15} \times \dfrac {14}{15} \times \dfrac {14}{15} = 0.813 \]

You have a slightly higher chance of not being picked compared to when she picked a new person for each question. However, you now may be picked more than once.

Exercise \(\PageIndex{2}\)

Under the setup of Example \(\PageIndex{3}\), what is the probability of being picked to answer all three questions?

**Solution**

P(not being picked on any of the three questions) = \(\dfrac { 1}{15}^3 = 0.00030\).

If we sample from a small population **without replacement**, we no longer have independence between our observations. In Example \(\PageIndex{2}\), the probability of not being picked for the second question was conditioned on the event that you were not picked for the first question. In Example \(\PageIndex{4}\), the professor sampled her students **with replacement**: she repeatedly sampled the entire class without regard to who she already picked.

Exercise \(\PageIndex{2}\)

Your department is holding a raffle. They sell 30 tickets and offer seven prizes.

- They place the tickets in a hat and draw one for each prize. The tickets are sampled without replacement, i.e. the selected tickets are not placed back in the hat. What is the probability of winning a prize if you buy one ticket?
- What if the tickets are sampled with replacement?

**Answer**

(a) First determine the probability of not winning. The tickets are sampled without replacement, which means the probability you do not win on the first draw is 29/30, 28/29 for the second, ..., and 23/24 for the seventh. The probability you win no prize is the product of these separate probabilities: 23/30. That is, the probability of winning a prize is 1 - 23/30 = 7/30 = 0.233.

(b) When the tickets are sampled with replacement, there are seven independent draws. Again we first nd the probability of not winning a prize: \(\dfrac {29}{30}^7 = 0.789\). Thus, the probability of winning (at least) one prize when drawing with replacement is 0.211.

Exercise \(\PageIndex{3}\)

Compare your answers in Exercise \(\PageIndex{2}\). How much influence does the sampling method have on your chances of winning a prize?

**Answer**

There is about a 10% larger chance of winning a prize when using sampling without replacement. However, at most one prize may be won under this sampling procedure.

Had we repeated Exercise \(\PageIndex{2}\) with 300 tickets instead of 30, we would have found something interesting: the results would be nearly identical. The probability would be 0.0233 without replacement and 0.0231 with replacement. When the sample size is only a small fraction of the population (under 10%), observations are nearly independent even when sampling without replacement.

### Contributors

- David M Diez (Google/YouTube)
- Christopher D Barr (Harvard School of Public Health)
- Mine Çetinkaya-Rundel (Duke University)