Discussion
Let N = 4. Two units are selected for inclusion in the sample based on the following design. The \(i\)-th unit is selected with probability \(p_{i}\) where \(p_{1} = 1/6\), \(p_{2} = 1/6\), \(p_{3} = 2/3\). The second unit is selected randomly from among the remaining three units in the population. Then, the first unit is included in the sample with probability :
Answer: B
The probability is 4/9.
Since \(p_{1} + p_{2} + p_{3} = 1\), the fourth unit has \(p_{4} = 0\). Unit 1 enters the sample either on the first draw, or on the second draw after another unit is drawn first:
\[ \begin{aligned} P &= p_{1} + (1 - p_{1}) \times \frac{1}{3} \\ &= \frac{1}{6} + \frac{5}{6} \times \frac{1}{3} = \frac{3}{18} + \frac{5}{18} = \frac{4}{9} \end{aligned} \]
2/9 forgets the first draw, and 6/9 and 7/9 overcount the second.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development