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Sampling without Replacement — Definition, Formula & Examples

Sampling without replacement is a way of choosing a sample where each item you pick is set aside and cannot be picked again. Every member of the population appears in the sample at most once.

A sampling procedure in which a selected unit is removed from the population before the next selection. The draws are therefore dependent: if a population of size NN contains KK successes, the probability of a success on a later draw depends on the outcomes of earlier draws. For example, P(2nd success∣1st success)=K−1N−1P(\text{2nd success} \mid \text{1st success}) = \frac{K-1}{N-1}.

Key Formula

P(first two draws are both successes)=KN⋅K−1N−1P(\text{first two draws are both successes}) = \frac{K}{N}\cdot\frac{K-1}{N-1}
Where:
  • NN = Population size before any draws
  • KK = Number of successes in the population before any draws
  • K−1N−1\frac{K-1}{N-1} = Probability of a success on draw 2, given draw 1 was a success. One success and one item have been removed.

Worked Example

Problem: A club has 10 members: 4 juniors and 6 seniors. You randomly choose 2 members without replacement to attend a meeting. What is the probability that both are juniors?
Step 1: Find the probability that the first person chosen is a junior. There are 4 juniors out of 10 members.
P(1st junior)=410=25P(\text{1st junior}) = \frac{4}{10} = \frac{2}{5}
Step 2: One junior is now removed. That leaves 3 juniors among 9 members.
P(2nd junior∣1st junior)=39=13P(\text{2nd junior} \mid \text{1st junior}) = \frac{3}{9} = \frac{1}{3}
Step 3: Multiply the two probabilities using the general multiplication rule.
410⋅39=1290=215≈0.133\frac{4}{10}\cdot\frac{3}{9} = \frac{12}{90} = \frac{2}{15} \approx 0.133
Answer: The probability is 215≈0.133\frac{2}{15} \approx 0.133. With replacement, it would be (0.4)2=0.16(0.4)^2 = 0.16. Removing the first junior lowers the chance of a second one.

Why It Matters

Real surveys and polls sample without replacement, because nobody is interviewed twice. In AP Statistics, the binomial model and the formula σp^=p(1−p)n\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} both assume independent draws. That assumption is only reasonable when the sample is at most 10% of the population, which is the 10% condition you check on the AP exam.

Common Mistakes

Mistake: Using the same probability on every draw, such as 410⋅410\frac{4}{10}\cdot\frac{4}{10}, when items are not replaced.
Correction: After each draw, subtract the chosen item from both the count of successes and the total. Use 410⋅39\frac{4}{10}\cdot\frac{3}{9}. You may treat the draws as approximately independent only when the 10% condition holds.

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