Skip to main content

Introductory Statistics

Section 3.3 Sampling from a small population

When we sample observations from a population, usually we’re only sampling a small fraction of the possible individuals or cases. However, sometimes our sample size is large enough or the population is small enough that we sample more than 10% of a population without replacement (meaning we do not have a chance of sampling the same cases twice). Sampling such a notable fraction of a population can be important for how we analyze the sample.

Example 3.3.1. Selecting a student at random.

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 \(1/15\text{,}\) or about 0.067.

Example 3.3.2. Three questions without replacement.

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 \(14/15\text{.}\) 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 \(13/14\text{.}\) Similarly, the probability you are again not picked on the third question is \(12/13\text{,}\) and the probability of not being picked for any of the three questions is
\begin{align*} P(\text{not picked in 3 questions}) \amp= P(\text{Q1} = \text{not picked, Q2} = \text{not picked, Q3} = \text{not picked})\\ \amp= \frac{14}{15} \times \frac{13}{14} \times \frac{12}{13} = \frac{12}{15} = 0.80 \end{align*}

Checkpoint 3.3.3.

What rule permitted us to multiply the probabilities in ExampleΒ 3.3.2?
Solution.
The three probabilities we computed were actually one marginal probability, \(P(\text{Q1} = \text{not picked})\text{,}\) and two conditional probabilities:
\begin{gather*} P(\text{Q2} = \text{not picked} \mid \text{Q1} = \text{not picked})\\ P(\text{Q3} = \text{not picked} \mid \text{Q1} = \text{not picked, Q2} = \text{not picked}) \end{gather*}
Using the General Multiplication Rule, the product of these three probabilities is the probability of not being picked in 3 questions.

Example 3.3.4. Three questions with replacement.

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 \(14/15\text{.}\) Thus, we can use the Multiplication Rule for independent processes.
\begin{align*} P(\text{not picked in 3 questions}) \amp= P(\text{Q1} = \text{not picked, Q2} = \text{not picked, Q3} = \text{not picked})\\ \amp= \frac{14}{15} \times \frac{14}{15} \times \frac{14}{15} = 0.813 \end{align*}
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.

Checkpoint 3.3.5.

Under the setup of ExampleΒ 3.3.4, what is the probability of being picked to answer all three questions?
Solution.
\(P(\text{being picked to answer all three questions}) = \left(\frac{1}{15}\right)^3 = 0.00030\)
If we sample from a small population without replacement, we no longer have independence between our observations. In ExampleΒ 3.3.2, 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 ExampleΒ 3.3.4, the professor sampled her students with replacement: she repeatedly sampled the entire class without regard to who she already picked.

Checkpoint 3.3.6.

Your department is holding a raffle. They sell 30 tickets and offer seven prizes.
  1. 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?
  2. What if the tickets are sampled with replacement?
Solution.
(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 \(29/30\text{,}\) \(28/29\) for the second, \(\ldots\text{,}\) and \(23/24\) for the seventh. The probability you win no prize is the product of these separate probabilities: \(23/30\text{.}\) That is, the probability of winning a prize is \(1 - 23/30 = 7/30 = 0.233\text{.}\)
(b) When the tickets are sampled with replacement, there are seven independent draws. Again we first find the probability of not winning a prize: \((29/30)^7 = 0.789\text{.}\) Thus, the probability of winning (at least) one prize when drawing with replacement is 0.211.

Checkpoint 3.3.7.

Compare your answers in CheckpointΒ 3.3.6. How much influence does the sampling method have on your chances of winning a prize?
Solution.
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 CheckpointΒ 3.3.6 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.

Exercises Exercises

1. Marbles in an urn.

Imagine you have an urn containing 5 red, 3 blue, and 2 orange marbles in it.
  1. What is the probability that the first marble you draw is blue?
  2. Suppose you drew a blue marble in the first draw. If drawing with replacement, what is the probability of drawing a blue marble in the second draw?
  3. Suppose you instead drew an orange marble in the first draw. If drawing with replacement, what is the probability of drawing a blue marble in the second draw?
  4. If drawing with replacement, what is the probability of drawing two blue marbles in a row?
  5. When drawing with replacement, are the draws independent? Explain.

3. Chips in a bag.

Imagine you have a bag containing 5 red, 3 blue, and 2 orange chips.
  1. Suppose you draw a chip and it is blue. If drawing without replacement, what is the probability the next is also blue?
  2. Suppose you draw a chip and it is orange, and then you draw a second chip without replacement. What is the probability this second chip is blue?
  3. If drawing without replacement, what is the probability of drawing two blue chips in a row?
  4. When drawing without replacement, are the draws independent? Explain.

4. Books on a bookshelf.

The table below shows the distribution of books on a bookcase based on whether they are nonfiction or fiction and hardcover or paperback.
Table 3.3.8. Books by type and format
Hardcover Paperback Total
Fiction 13 59 72
Nonfiction 15 8 23
Total 28 67 95
  1. Find the probability of drawing a hardcover book first then a paperback fiction book second when drawing without replacement.
  2. Determine the probability of drawing a fiction book first and then a hardcover book second, when drawing without replacement.
  3. Calculate the probability of the scenario in part (b), except this time complete the calculations under the scenario where the first book is placed back on the bookcase before randomly drawing the second book.
  4. The final answers to parts (b) and (c) are very similar. Explain why this is the case.

5. Student outfits.

In a classroom with 24 students, 7 students are wearing jeans, 4 are wearing shorts, 8 are wearing skirts, and the rest are wearing leggings. If we randomly select 3 students without replacement, what is the probability that one of the selected students is wearing leggings and the other two are wearing jeans? Note that these are mutually exclusive clothing options.

6. The birthday problem.

Suppose we pick three people at random. For each of the following questions, ignore the special case where someone might be born on February 29th, and assume that births are evenly distributed throughout the year.
  1. What is the probability that the first two people share a birthday?
  2. What is the probability that at least two people share a birthday?