ISS PREVIOUS YEAR 2016 PAPER-1 SET-A Q.no.- 10
- Shivani Rana
- 2 days ago
- 1 min read
ISS PREVIOUS YEAR 2016 PAPER-1 SET-A
![For the distribution:
f(x) = [1 / B(p, q)] × [x^(p − 1) / (1 + x)^(p + q)],where 0 < x < ∞, p > 0, q > 0,
find the harmonic mean.
Options:
(a) p / (p + q) (b) 1 / p (c) (p − 1) / q (d) (p + 1) / (q − 1)](https://static.wixstatic.com/media/8ffd4d_929597f8730a4e7fb508b389d8141e98~mv2.jpg/v1/fill/w_400,h_400,al_c,q_80,enc_avif,quality_auto/8ffd4d_929597f8730a4e7fb508b389d8141e98~mv2.jpg)
Expected Number of Unique Elements in a Sample
Question
A simple random sample of size 10 is selected with replacement from a population of size 100. What is the expected number of unique elements in the sample?
Options:(a) 100 × (99/100)¹⁰(b) 100 × [1 − (99/100)¹⁰](c) 100 × (9/10)¹⁰(d) 100 × (1 − 99/100)¹⁰
Solution
This problem is based on the concept of expected number of distinct (unique) elements.
Consider any one element from the population.
The probability that this element is not selected in one draw is:
99/100
Since sampling is done with replacement, the draws are independent.
So, the probability that the element is never selected in 10 draws is:
(99/100)¹⁰
Therefore, the probability that the element is selected at least once is:
1 − (99/100)¹⁰
Now, there are 100 such elements in the population.
so, the expected number of elements that appear at least once in the sample is:
Expected number of unique elements =100 × [1 − (99/100)¹⁰]
Final Answer
Option (b):100 × [1 − (99/100)¹⁰]
ISS PREVIOUS YEAR 2016 PAPER-1 SET-A

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