Discussion
A probability proportional to size (PPS) without replacement sample of size 2 is to be drawn from a population of N = 3 units. Let \(z_{1} = 1/4\), \(z_{2} = 1/4\), \(z_{3} = 1/2\). The \(i\)-th unit is selected with probability \(z_{i}\), \(i = 1, 2, 3\). If the first unit selected is \(i\), then the \(j\)-th unit is selected for inclusion in the sample with probability \(z_{j}/(1 - z_{i})\), \(j \ne i\). Then, the probability that the third unit is the population is included in the sample is given by :
Answer: C
The probability is 5/6.
Unit 3 is in the sample if it is drawn first, or if unit 1 or unit 2 is drawn first and unit 3 second:
\[ \begin{aligned} \pi_{3} &= z_{3} + z_{1}\frac{z_{3}}{1 - z_{1}} + z_{2}\frac{z_{3}}{1 - z_{2}} \\ &= \tfrac{1}{2} + 2 \cdot \tfrac{1}{4}\cdot\tfrac{1/2}{3/4} \\ &= \tfrac{1}{2} + \tfrac{1}{3} = \tfrac{5}{6} \end{aligned} \]
The paper's 'the third unit is the population' means 'in the population'. 1/2 is only the first-draw chance; 2/3 is the inclusion probability with equal probabilities; and 1 would hold only if unit 3 could not be missed. As a check, \(\pi_{1} = \pi_{2} = 7/12\), and \(7/12 + 7/12 + 5/6 = 2\), the sample size.
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