Skip to content
Part CCSIR NET December 2024two-draws-cannot-contain-three-units

Two draws cannot contain three units

Consider a probability proportional to size without replacement sample involving two draws from a population of N (> 6) units with normed size measures pi′s(pi>0,i=1,2,…,N;∑(i=1..N)pi=1)p_{i}'s (p_{i} > 0, i = 1, 2, \dots, N; \sum(i=1..N)p_{i} = 1). Then, which of the following statements are true?

  1. A.P(Unit 1 is included in the sample)=p1) = p_{1}
  2. B.P(Unit 1 and Unit 3 are included in the sample)=p1p3[1/(1−p1)+1/(1−p3)]) = p_{1}p_{3}[1/(1 - p_{1}) + 1/(1 - p_{3})]✓ Correct
  3. C.P(Unit 1, Unit 3 and Unit 5 are included in the sample)=p1p3p5[1/((1−p1)(1−p3))+1/((1−p3)(1−p5))+1/((1−p5)(1−p1))]) = p_{1}p_{3}p_{5}[1/((1 - p_{1})(1 - p_{3})) + 1/((1 - p_{3})(1 - p_{5})) + 1/((1 - p_{5})(1 - p_{1}))]
  4. D.Expected number of distinct units in the sample is 2[1−∑(i=1..N)(1−pi)2]2[1 - \sum(i=1..N)(1 - p_{i})^{2}]

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests execution slip.

See pricing

50 are analysed free — try those first.

The trap it tests

Execution slip

The idea was right. The computation was not.

Drill statements like this

Related counterexample: Systematic sampling is always at least as efficient as SRS

More on this topic

The chapter behind this: Sampling designs — free to read

From Sampling and Design of Experiments › SRS, stratified and systematic sampling

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

ShareWhatsAppTelegram