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Exact sampling distribution of a sample proportion

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A population has 3 successes and 2 failures. Two units are selected without replacement. Which line correctly gives the probability of exactly 1 success?

Year 12, VCE Units 3 & 4, Chapter 14, 4 difficulty levels, No calculatorBuild a worksheet

An exact finite sampling distribution

When a sample is selected without replacement, count each unordered sample exactly. For each possible success count, combine the success choices and failure choices, then divide by the total number of samples

\[\Pr(X=x)=\frac{\binom{A}{x}\binom{B}{n-x}}{\binom{A+B}{n}}\]
  1. Use combinations without replacement A finite sample is an unordered group, so use combinations rather than independent powers \(\displaystyle \text{without replacement}\Rightarrow\binom{N}{n}\)
  2. Find feasible success counts The number of successes must allow enough successes and failures to fill the sample \(\displaystyle A=3,\quad B=2,\quad n=2\Rightarrow x=0,1,2\)
  3. Count each row Multiply the number of ways to choose the successes by the number of ways to choose the failures \(\displaystyle c_1=\binom{3}{1}\binom{2}{1}=3\times2=6\)
  4. Divide by the total sample space Use the combination for all population members and check that all row probabilities sum to one \(\displaystyle T=\binom{5}{2}=10,\quad\Pr(X=1)=\frac{6}{10}=\frac{3}{5}\)

Worked example

A finite population contains 6 success-labelled units and 4 failure-labelled units. Two units are selected without replacement. Find the probability that exactly one selected unit is a success

\[\Pr(X=1)=\frac{\binom{6}{1}\binom{4}{1}}{\binom{10}{2}}\]
  1. \(\displaystyle N=6+4=10\)The population has 10 units altogether
  2. \(\displaystyle c_1=\binom{6}{1}\binom{4}{1}=6\times4=24\)Choose one success and one failure for the required sample
  3. \(\displaystyle T=\binom{10}{2}=45\)Choose any two units to count the full unordered sample space
  4. \(\displaystyle \Pr(X=1)=\frac{24}{45}=\frac{8}{15}\)Divide the favourable sample count by the total sample count and reduce

Common mistakes

Using equal row probabilities
✕\(\displaystyle \Pr(X=0)=\Pr(X=1)=\Pr(X=2)=\frac{1}{3}\)✓\(\displaystyle \Pr(X=1)=\frac{\binom{3}{1}\binom{2}{1}}{\binom{5}{2}}=\frac{3}{5}\)
The combination count for a row determines its probability, so feasible rows need not be equally likely
Treating draws as independent
✕\(\displaystyle \Pr(X=1)=\binom{2}{1}\left(\frac{3}{5}\right)\left(\frac{2}{5}\right)\)✓\(\displaystyle \Pr(X=1)=\frac{\binom{3}{1}\binom{2}{1}}{\binom{5}{2}}\)
Without replacement changes the available population after each selection, so the draws are dependent
Using the wrong denominator
✕\(\displaystyle \Pr(X=1)=\frac{\binom{3}{1}\binom{2}{1}}{3}=\frac{6}{3}\)✓\(\displaystyle \Pr(X=1)=\frac{\binom{3}{1}\binom{2}{1}}{\binom{5}{2}}=\frac{3}{5}\)
The denominator must count every unordered sample of the stated size

Difficulty levels

introComplete one missing probability in a small exact distribution
standardComplete selected rows in a finite sampling-distribution table
challengeFind an event or conditional probability by combining table entries
examSelect the without-replacement model or total sample count, then calculate an exact probability

Questions about this topic

Why are combinations used for a finite sample?+

The order of the selected units is not recorded, so each unordered group is counted once with a combination

Why do the probabilities in the rows add to one?+

The feasible success-count rows divide all possible samples into disjoint cases, so their exact probabilities cover the whole sample space

How is a conditional probability found from the table?+

Add the probabilities for the rows satisfying both conditions and divide by the total probability for the given condition

Can I use a calculator?+

No. This topic is practised without a calculator

Which year level and chapter is this?+

Year 12, VCE Units 3 & 4, Mathematical Methods, chapter 14: Statistical inference

In this chapter

Populations, samples, parameters, statistics and sampling methodsAllocate a stratified random sampleFind a percentage of a quantityCalculate a sample proportion and list its possible valuesExact sampling distribution of a sample proportionFinite events, probability assignments and complementsConditional probability from exact representationsEvaluate factorial and combination expressionsConvert between fractions, decimals and percentagesLarge-sample conditions and the sample-proportion distributionCalculate a standard score or actual valueUse z-score boundaries to estimate normal-model percentages and countsCalculate the standard deviation from the formulaSolve a linear equationSolve a quadratic exactly with the quadratic formulaApproximate confidence intervals for a population proportionRearrange literal equations and recognise equivalent line formsInterpret simulated sampling distributions and interval variationDesign and interpret a probability simulation