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 $p_{i}^{'} s (p_{i} > 0, i = 1, 2, \dots, N; \sum (i = 1.. N) p_{i} = 1)$ . Then, which of the following statements are true?

A.P(Unit 1 is included in the sample $) = p_{1}$
B.P(Unit 1 and Unit 3 are included in the sample $) = p_{1} p_{3} [1/ (1 - p_{1}) + 1/ (1 - p_{3})]$ ✓
C.P(Unit 1, Unit 3 and Unit 5 are included in the sample $) = p_{1} p_{3} p_{5} [1/ ((1 - p_{1}) (1 - p_{3})) + 1/ ((1 - p_{3}) (1 - p_{5})) + 1/ ((1 - p_{5}) (1 - p_{1}))]$
D.Expected number of distinct units in the sample is $2 [1 - \sum (i = 1.. N) (1 - p_{i})^{2}]$

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.

See pricing

50 are analysed free — try those first.

Two draws cannot contain three units

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