This solution calculates the probability that a specific unit (the 4th) is included in a simple random sample (SRS) of size $n$ drawn without replacement from a population of size $N$, while another specific unit (the 6th) is excluded.
We use the concept of conditional probability for sampling without replacement.
Define Events:
Probability of 4th Unit Inclusion:
In an SRS of size $n$ from a population of size $N$, the probability that any specific unit is included is $\frac{n}{N}$.
$ P(E_{4, \text{in}}) = \frac{n}{N} $Conditional Probability of 6th Unit Exclusion:
Given that the 4th unit is already included in the sample, there are $n-1$ remaining spots to be filled from the remaining $N-1$ population units. The 6th unit is one of these $N-1$ units.
The probability that the 6th unit IS included among the remaining $n-1$ selections is $\frac{n-1}{N-1}$.
Therefore, the probability that the 6th unit is NOT included, given the 4th was included, is:
$ P(E_{6, \text{out}} | E_{4, \text{in}}) = 1 - P(\text{6th included} | E_{4, \text{in}}) = 1 - \frac{n-1}{N-1} $ $ P(E_{6, \text{out}} | E_{4, \text{in}}) = \frac{(N-1) - (n-1)}{N-1} = \frac{N-n}{N-1} $Combined Probability:
Using the multiplication rule for conditional probability:
$ P(E_{4, \text{in}} \cap E_{6, \text{out}}) = P(E_{4, \text{in}}) \times P(E_{6, \text{out}} | E_{4, \text{in}}) $ $ P(E_{4, \text{in}} \cap E_{6, \text{out}}) = \frac{n}{N} \times \frac{N-n}{N-1} $ $ P(E_{4, \text{in}} \cap E_{6, \text{out}}) = \frac{n(N-n)}{N(N-1)} $This result matches option B.
Suppose there are $k$ groups each consisting of $N$ boys. We want to estimate the mean age $\mu$ of these $kN$ boys. Fix $1 < n < N$ and consider the following two sampling schemes.
I. Draw a simple random sample without replacement of size $kn$ out of all $kN$ boys.
II. From each of the $k$ groups draw a simple random sample with replacement of size $n$.
Let $\bar{Y}$ and $\bar{Y}_G$ be the respective sample mean ages for the two schemes. Which of the following are true?